Topic 5 — tasks — Mathematik

1

Task One correct answer

Simplify \(\sqrt[3]{\sqrt[4]{a^6}}\) if \(a\ge 0\).

А \(\sqrt[3]{a^2}\) Б \(\sqrt a\) В \(a^2\) Г \(\sqrt[6]{a}\) Д \(\sqrt[7]{a^6}\)

Solution

Write the roots as powers: \(\sqrt[3]{\sqrt[4]{a^6}}=(a^6)^{\frac14\cdot\frac13}=a^{6/12}=a^{1/2}=\sqrt a\). Since \(a\ge0\), no absolute value is needed. Answer: B.

2

Task Numeric answer

Calculate the value of \(20\sqrt6-\left(\dfrac4{\sqrt2}+5\sqrt3\right)^2\).

Your answer

Solution

\(\dfrac4{\sqrt2}=2\sqrt2\). Then \((2\sqrt2+5\sqrt3)^2=8+20\sqrt6+75=83+20\sqrt6\). Hence \(20\sqrt6-(83+20\sqrt6)=-83\). Answer: \(-83\).

3

Task One correct answer

Simplify \(\dfrac5{a-9}:\dfrac1{2\sqrt a-6}\).

А \(\dfrac{10}{\sqrt a-3}\) Б \(\dfrac5{2\sqrt a+6}\) В \(\dfrac{\sqrt a+3}{10}\) Г \(\dfrac{10}{\sqrt a+3}\) Д \(\dfrac{2\sqrt a-6}{5}\)

Solution

Replace division by a fraction with multiplication by its reciprocal: \(\dfrac5{a-9}(2\sqrt a-6)=\dfrac{10(\sqrt a-3)}{a-9}\). Since \(a-9=(\sqrt a-3)(\sqrt a+3)\), this simplifies to \(\dfrac{10}{\sqrt a+3}\). Answer: D.

4

Task One correct answer

If \(F=\dfrac{GMm}{R^2}\) and \(R>0\), then \(R=\)

А \(\sqrt{FGMm}\) Б \(\sqrt{\dfrac{Mm}{GF}}\) В \(\sqrt{\dfrac{GF}{Mm}}\) Г \(\sqrt{\dfrac{F}{GMm}}\) Д \(\sqrt{\dfrac{GMm}{F}}\)

Solution

From \(F=\dfrac{GMm}{R^2}\), we get \(FR^2=GMm\), so \(R^2=\dfrac{GMm}{F}\). Since \(R>0\), take the positive root: \(R=\sqrt{\dfrac{GMm}{F}}\). Answer: E.

5

Task One correct answer

Simplify \((\sqrt2-a)(\sqrt2+a)\).

А \(2-a\) Б \(2-a^2\) В \(\sqrt2-a^2\) Г \(2-\sqrt a\) Д \(4\sqrt2-a^2\)

Solution

Using the difference of squares: \((\sqrt2-a)(\sqrt2+a)=(\sqrt2)^2-a^2=2-a^2\). Answer: B.

6

Task One correct answer

Evaluate \(|1-\sqrt3|\).

А \(-1-\sqrt3\) Б \(\sqrt3-1\) В \(1-\sqrt3\) Г \(1+\sqrt3\) Д \(1\)

Solution

Since \(\sqrt3>1\), we have \(1-\sqrt3<0\). Therefore \(|1-\sqrt3|=-(1-\sqrt3)=\sqrt3-1\). Answer: B.

7

Task Numeric answer

Calculate \(\dfrac{\sqrt{18-8\sqrt2}}{\sqrt2}\cdot\sqrt{9+4\sqrt2}\).

Your answer

Solution

\(18-8\sqrt2=(4-\sqrt2)^2\) and \(9+4\sqrt2=(1+2\sqrt2)^2\). Thus \(\dfrac{4-\sqrt2}{\sqrt2}(1+2\sqrt2)=(2\sqrt2-1)(1+2\sqrt2)=7\). Answer: \(7\).

8

Task One correct answer

Choose the interval containing the solution of \(\sqrt{x+12}=3\).

А \([-12;-6)\) Б \([-6;0)\) В \([0;6)\) Г \([6;12)\) Д \([12;+\infty)\)

Solution

Squaring gives \(x+12=9\), so \(x=-3\). The number \(-3\) lies in \([-6;0)\). Answer: B.

9

Task Numeric answer

Calculate \(\sqrt{9a^2-24a+16}-\sqrt[3]{27a^3}\) for \(a=0.7\).

Your answer

Solution

\(9a^2-24a+16=(3a-4)^2\), so \(\sqrt{9a^2-24a+16}=|3a-4|\), and \(\sqrt[3]{27a^3}=3a\). For \(a=0.7\): \(1.9-2.1=-0.2\). Answer: \(-0.2\).

10

Task Matching

Match each expression (1–3) with its value (A–E) if \(x=\sqrt5-1\).

А \(-1\) Б \(1\) В \(4\) Г \(5\) Д \(6\)

1. \(|x-\sqrt5|\) 2. \((\sqrt5+1)x\) 3. \(x^2+2x+1\)

Solution

1) \(|x-\sqrt5|=|-1|=1\) — B. 2) \((\sqrt5+1)(\sqrt5-1)=4\) — C. 3) \(x^2+2x+1=(x+1)^2=5\) — D. Matching: 1–B, 2–C, 3–D.

11

Task One correct answer

Simplify \(15\sqrt{\dfrac25}-\sqrt{160}\).

А \(\sqrt{10}\) Б \(-\sqrt{10}\) В \(11\) Г \(-2\sqrt5\) Д \(-37\sqrt2\)

Solution

\(15\sqrt{\dfrac25}=15\cdot\dfrac{\sqrt{10}}5=3\sqrt{10}\), and \(\sqrt{160}=4\sqrt{10}\). Hence \(3\sqrt{10}-4\sqrt{10}=-\sqrt{10}\). Answer: B.

12

Task One correct answer

Simplify \(\sqrt[3]{18z^4}\cdot\sqrt[3]{12z^5}\).

А \(6z^3\) Б \(18z^3\) В \(12\) Г \(\sqrt[3]{6z}\) Д \(6\)

Solution

Combine the cube roots: \(\sqrt[3]{18z^4}\cdot\sqrt[3]{12z^5}=\sqrt[3]{216z^9}=\sqrt[3]{(6z^3)^3}=6z^3\). Answer: A.

13

Task Numeric answer

Calculate \(\sqrt[5]{4}\cdot\sqrt[10]{64}\).

Your answer

Solution

\(\sqrt[5]{4}=\sqrt[5]{2^2}\), and \(\sqrt[10]{64}=\sqrt[10]{2^6}=\sqrt[5]{2^3}\). Thus \(\sqrt[5]{2^2}\cdot\sqrt[5]{2^3}=\sqrt[5]{2^5}=2\). Answer: \(2\).

14

Task One correct answer

Simplify \(\dfrac{\sqrt2}{\sqrt2-\sqrt7}-\dfrac{\sqrt7}{\sqrt2+\sqrt7}\).

А \(7\sqrt2\) Б \(2\sqrt7\) В \(\sqrt{14}\) Г \(-2{,}8\) Д \(-1{,}8\)

Solution

The common denominator is \((\sqrt2-\sqrt7)(\sqrt2+\sqrt7)=2-7=-5\). The numerator is \(2+\sqrt{14}-\sqrt{14}+7=9\). Hence \(9/(-5)=-1.8\). Answer: E.

15

Task One correct answer

Arrange \(\sqrt[3]{2},\ 1,\ \sqrt[5]{3}\) in increasing order.

А \(1,\ \sqrt[3]{2},\ \sqrt[5]{3}\) Б \(1,\ \sqrt[5]{3},\ \sqrt[3]{2}\) В \(\sqrt[3]{2},\ \sqrt[5]{3},\ 1\) Г \(\sqrt[5]{3},\ 1,\ \sqrt[3]{2}\) Д \(\sqrt[3]{2},\ 1,\ \sqrt[5]{3}\)

Solution

All numbers are positive. Compare their 15th powers: \((\sqrt[3]{2})^{15}=32\), \((\sqrt[5]{3})^{15}=27\), and \(1^{15}=1\). Therefore \(1<\sqrt[5]{3}<\sqrt[3]{2}\). Answer: B.

16

Task Numeric answer

Calculate \(\dfrac{3\sqrt2-5}{\sqrt2-1}+\dfrac{\sqrt{24}-\sqrt{300}}{\sqrt3}\).

Your answer

Solution

Rationalizing the first fraction gives \(\dfrac{3\sqrt2-5}{\sqrt2-1}=1-2\sqrt2\). Also, \(\dfrac{\sqrt{24}-\sqrt{300}}{\sqrt3}=2\sqrt2-10\). Their sum is \(-9\). Answer: \(-9\).

17

Task One correct answer

Which interval contains \(\sqrt[4]{30}\)?

А \((1;2)\) Б \((2;3)\) В \((3;4)\) Г \((4;5)\) Д \((5;6)\)

Solution

Since \(2^4=16<30<81=3^4\), we get \(2<\sqrt[4]{30}<3\). Answer: B.

18

Task One correct answer

Among \(a=\sqrt5-2\), \(b=2\sqrt3-3\sqrt2\), \(c=\sqrt[3]{3}-\sqrt[3]{2}\), choose all positive numbers.

А \(a\) Б \(c\) В \(a;\ b\) Г \(a;\ c\) Д \(a;\ b;\ c\)

Solution

\(a>0\) because \(\sqrt5>2\). For \(b\), \((2\sqrt3)^2=12<18=(3\sqrt2)^2\), so \(b<0\). Also \(c>0\) because \(\sqrt[3]{3}>\sqrt[3]{2}\). Thus the positive numbers are \(a\) and \(c\). Answer: D.

19

Task Numeric answer

Calculate \((\sqrt[6]{27}-\sqrt[4]{100})(\sqrt[6]{27}+\sqrt[4]{100})\).

Your answer

Solution

Use the difference of squares: \((u-v)(u+v)=u^2-v^2\). Thus \((\sqrt[6]{27})^2-(\sqrt[4]{100})^2=\sqrt[3]{27}-\sqrt{100}=3-10=-7\). Answer: \(-7\).

20

Task One correct answer

Simplify \(\dfrac{2\sqrt2+1}{\sqrt2+1}\).

А \(2\) Б \(\sqrt2+1\) В \(3+\sqrt2\) Г \(3+2\sqrt2\) Д \(3-\sqrt2\)

Solution

Multiply numerator and denominator by \(\sqrt2-1\): \(\dfrac{(2\sqrt2+1)(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}=3-\sqrt2\). Answer: E.

21

Task One correct answer

Which of the following numbers is irrational?

А \(\sqrt{0{,}64}\) Б \(\sqrt{20}\) В \(\pi^0\) Г \(\sqrt[3]{8}\) Д \(2{,}7\)

Solution

\(\sqrt{0{,}64}=0.8\), \(\pi^0=1\), \(\sqrt[3]{8}=2\), and \(2.7\) are rational. \(\sqrt{20}=2\sqrt5\) is irrational. Answer: B.

22

Task Matching

Match the numerical expressions (1–4) with their values (A–E).

А \(15\) Б \(6\) В \(5\) Г \(3\) Д \(1\)

1. \((\sqrt3+\sqrt2)^2-2\sqrt6\) 2. \((\sqrt5-2)(\sqrt5+2)\) 3. \((3\sqrt3-\sqrt{12})^2\) 4. \(\sqrt2(\sqrt{50}-\sqrt8)\)

Solution

1) \((\sqrt3+\sqrt2)^2-2\sqrt6=5\) — C. 2) \((\sqrt5-2)(\sqrt5+2)=1\) — E. 3) Since \(\sqrt{12}=2\sqrt3\), \((3\sqrt3-\sqrt{12})^2=3\) — D. 4) \(\sqrt2(\sqrt{50}-\sqrt8)=\sqrt2(5\sqrt2-2\sqrt2)=6\) — B. Matching: 1–C, 2–E, 3–D, 4–B.

23

Task One correct answer

Which of the following numbers is rational?

А \(\sqrt[3]{9}\) Б \(\sqrt{10}\) В \(\pi\) Г \(\sqrt{3{,}6}\) Д \(\sqrt{0{,}64}\)

Solution

\(\sqrt[3]{9}\), \(\sqrt{10}\), \(\pi\), and \(\sqrt{3.6}\) are irrational. \(\sqrt{0.64}=0.8=\dfrac45\) is rational. Answer: E.

24

Task One correct answer

Calculate \(\sqrt[4]{16\cdot81}\).

А \(6\) Б \(12\) В \(18\) Г \(36\) Д \(72\)

Solution

\(16=2^4\) and \(81=3^4\). Hence \(\sqrt[4]{16\cdot81}=\sqrt[4]{6^4}=6\). Answer: A.

25

Task One correct answer

Simplify \(\dfrac{\sqrt[3]{64}}{64}\).

А \(\dfrac1{16}\) Б \(\dfrac14\) В \(\dfrac13\) Г \(4\) Д \(16\)

Solution

Since \(64=4^3\), \(\sqrt[3]{64}=4\). Therefore \(\dfrac{\sqrt[3]{64}}{64}=\dfrac4{64}=\dfrac1{16}\). Answer: A.

26

Task One correct answer

Calculate \(\sqrt{(-2)^2}+\sqrt[3]{(-3)^3}\).

А \(-23\) Б \(-5\) В \(-1\) Г \(1\) Д \(5\)

Solution

For the square root, \(\sqrt{a^2}=|a|\), so \(\sqrt{(-2)^2}=2\). The cube root preserves the sign: \(\sqrt[3]{(-3)^3}=-3\). Thus \(2-3=-1\). Answer: C.

27

Task One correct answer

Calculate \(\dfrac{\sqrt[3]{189}}{\sqrt[3]{7}}\).

А \(3\) Б \(7\) В \(9\) Г \(21\) Д \(27\)

Solution

Use the quotient rule for cube roots: \(\dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}=\sqrt[3]{\dfrac ab}\). Thus \(\dfrac{\sqrt[3]{189}}{\sqrt[3]{7}}=\sqrt[3]{27}=3\). Answer: A.

28

Task One correct answer

Simplify \(\dfrac{\sqrt{12}\cdot\sqrt2}{\sqrt6}\).

А \(4\) Б \(3\) В \(2\) Г \(2\sqrt2\) Д \(\sqrt{\dfrac73}\)

Solution

Since \(\sqrt{12}=\sqrt2\sqrt6\), \(\dfrac{\sqrt{12}\sqrt2}{\sqrt6}=\dfrac{\sqrt2\sqrt6\sqrt2}{\sqrt6}=\sqrt2\sqrt2=2\). Answer: C.

29

Task One correct answer

Calculate \(\sqrt[3]{2^3}\cdot\sqrt{36}\).

А \(12\) Б \(6\sqrt[9]{2}\) В \(6\sqrt[6]{2}\) Г \(72\) Д \(6\sqrt2\)

Solution

We have \(\sqrt[3]{2^3}=2\) and \(\sqrt{36}=6\). Therefore \(\sqrt[3]{2^3}\cdot\sqrt{36}=2\cdot6=12\). Answer: A.

30

Task One correct answer

Calculate the value of \(2\sqrt{m+m+m}\) if \(m=\dfrac1{27}\).

А \(\dfrac23\) Б \(6\) В \(\dfrac16\) Г \(\dfrac32\) Д \(\dfrac29\)

Solution

Since \(m+m+m=3m\), for \(m=\dfrac1{27}\) we get \(2\sqrt{3m}=2\sqrt{\dfrac3{27}}=2\sqrt{\dfrac19}=2\cdot\dfrac13=\dfrac23\). Answer: A.

31

Task One correct answer

Simplify \( (\sqrt3-1)(1+\sqrt3) \).

А \(1\) Б \(2\) В \(\sqrt3\) Г \(2\sqrt3\) Д \(4\)

Solution

These are conjugate factors: \((\sqrt3-1)(\sqrt3+1)=(\sqrt3)^2-1^2=3-1=2\). Answer: B.

32

Task One correct answer

Calculate \((-3\sqrt3)^2\).

А \(-27\) Б \(-9\) В \(9\) Г \(18\) Д \(27\)

Solution

\((-3\sqrt3)^2=(-3)^2(\sqrt3)^2=9\cdot3=27\). Answer: E.

33

Task One correct answer

Simplify \(a^6\sqrt{a^4}\).

А \(a^8\) Б \(a^{10}\) В \(a^6\) Г \(|a|^8\) Д \(a^{12}\)

Solution

Since \(a^2\ge0\), \(\sqrt{a^4}=\sqrt{(a^2)^2}=a^2\). Hence \(a^6\sqrt{a^4}=a^6\cdot a^2=a^8\). Answer: A.

34

Task One correct answer

Simplify \(\dfrac{\sqrt9-\sqrt4}{\sqrt5}\).

А \(\sqrt5\) Б \(\dfrac{\sqrt5}{5}\) В \(\dfrac15\) Г \(5\) Д \(\dfrac1{25}\)

Solution

\(\sqrt9-\sqrt4=3-2=1\), so \(\dfrac1{\sqrt5}=\dfrac{\sqrt5}{5}\). Answer: B.

35

Task Matching

Match each numerical expression (1–3) with its value (A–E).

А \(-3\) Б \(\dfrac15\) В \(4\) Г \(3\) Д \(5\)

1. \(\sqrt[3]{-27}\) 2. \(\dfrac{\sqrt{12}}{\sqrt{300}}\) 3. \((\sqrt3-1)^2+2\sqrt3\)

Solution

1) \(\sqrt[3]{-27}=-3\) — A. 2) \(\dfrac{\sqrt{12}}{\sqrt{300}}=\sqrt{\dfrac1{25}}=\dfrac15\) — B. 3) \((\sqrt3-1)^2+2\sqrt3=4\) — C. Matching: 1–A, 2–B, 3–C.

36

Task One correct answer

Calculate \(\dfrac{\sqrt{9+a^2-6a}}{a-3}\) if \(a=2.5\).

А \(-1\) Б \(-\dfrac12\) В \(0\) Г \(\dfrac12\) Д \(1\)

Solution

Since \(9+a^2-6a=(a-3)^2\), we have \(\sqrt{(a-3)^2}=|a-3|\). For \(a=2.5\), \(|a-3|=0.5\) and \(a-3=-0.5\). Thus the value is \(-1\). Answer: A.

37

Task One correct answer

Choose the correct order of \(5\sqrt2\), \(7\), and \(\sqrt{51}\).

А \(5\sqrt2<7<\sqrt{51}\) Б \(7<5\sqrt2<\sqrt{51}\) В \(7<\sqrt{51}<5\sqrt2\) Г \(\sqrt{51}<7<5\sqrt2\) Д \(5\sqrt2<\sqrt{51}<7\)

Solution

All numbers are positive. Compare their squares: \(7^2=49\), \((5\sqrt2)^2=50\), \((\sqrt{51})^2=51\). Hence \(7<5\sqrt2<\sqrt{51}\). Answer: B.

38

Task One correct answer

Write \(\dfrac{a}{\sqrt[7]{a}}\) as a power with base \(a\), where \(a>0\).

А \(a^{-1/7}\) Б \(a^{1/7}\) В \(a^{6/7}\) Г \(a^7\) Д \(a^6\)

Solution

\(\sqrt[7]{a}=a^{1/7}\), so \(\dfrac{a}{\sqrt[7]{a}}=a^{1-1/7}=a^{6/7}\). Answer: C.

39

Task One correct answer

Rationalize the denominator and simplify \(\dfrac1{\sqrt{10}-3}\).

А \(\sqrt{10}-3\) Б \(3-\sqrt{10}\) В \(\sqrt{10}+3\) Г \(\dfrac{\sqrt{10}+3}{19}\) Д \(1\)

Solution

Multiply numerator and denominator by the conjugate \(\sqrt{10}+3\): \(\dfrac1{\sqrt{10}-3}=\dfrac{\sqrt{10}+3}{10-9}=\sqrt{10}+3\). Answer: C.

40

Task One correct answer

Simplify \(\dfrac{2\sqrt5-10}{\sqrt5}\).

А \(2-2\sqrt5\) Б \(2-\sqrt5\) В \(2\sqrt5-2\) Г \(-2\sqrt5\) Д \(2\)

Solution

Divide each term in the numerator by \(\sqrt5\): \(\dfrac{2\sqrt5}{\sqrt5}-\dfrac{10}{\sqrt5}=2-2\sqrt5\). Answer: A.

Irrational expressions