Homealgebra Grade 10 - Exponents and Radicals (Multiple Choice Questions) သူရိန်မင်း Wednesday, January 20, 2021 4 ဖြေကြည့်ပါ...၊ ရမှတ်နဲ့ အဖြေမှန်ကို ဖော်ပြပေးပါလိမ့်မယ်...။ မည်သည့်တွက်ချက်မှုဆိုင်ရာ ပစ္စည်းကိုမျှ အသုံးပြုခွင့် မရှိပါ။ မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။ 1. If $\displaystyle \left( \frac{a}{b} \right)^{x-1} =\left( \frac{b}{a} \right)^{x-3}$ , then the value of $\displaystyle x$ is A. $ \dfrac{1}{2}$ B. $2$ C. $\dfrac{7}{2}$ D. $1$ Explanation $\begin{aligned} \left(\frac{a}{b}\right)^{x-1} &=\left(\frac{b}{a}\right)^{x-3} \\\\ \left(\frac{a}{b}\right)^{x-1} &=\left(\left(\frac{a}{b}\right)^{-1}\right)^{x-3} \\\\ \left(\frac{a}{b}\right)^{x-1} &=\left(\frac{a}{b}\right)^{3-x} \\\\ \therefore x-1 &=3-x \\\\ 2 x &=4 \\\\ x &=2 \end{aligned}$ 2. If $\displaystyle 3^{x-y} =27$ and $\displaystyle 3^{x+y} =243$, then $x=$. A. $0$ B. $2$ C. $4$ D. $6$ Explanation $\begin{aligned} 3^{x-y} &=27 \\\\ 3^{x-y} &=3^{3} \\\\ x-y &=3 \ldots(1) \\\\ 3^{x+y} &=2 y^{3} \\\\ 3^{x+y} &=3^{5} \\\\ x+y &=5 \ldots(2)\\\\ \end{aligned}$ Solving equations $(1)$ and $(2)$, $\begin{aligned} &\\ x&=4 \end{aligned}$ 3. $\displaystyle \frac{243^{\frac{n}{5}}\times \:3^{2n+1}}{9^n\times 3^{n-1}}=$ A. $9$ B. $3^n$ C. $3$ D. $1$ Explanation $\begin{aligned} & \frac{243^{\frac{n}{5}} \times 3^{2 n+1}}{9^{n} \times 3^{n-1}} \\\\ =& \frac{\left(3^{5}\right)^{\frac{n}{5}} \times 3^{2 n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}} \\\\ =& \frac{3^{n} \times 3^{2 n+1}}{3^{2 n} \times 3^{n-1}} \\\\ =& \frac{3^{3 n+1}}{3^{3 n-1}} \\\\ =& 3^{3 n+1-3 n+1}\\\\ =&3^{2}\\\\ =&9 \end{aligned}$ 4. $\left(\displaystyle \frac{x^{b}}{x^{c}}\right)^{(b+c-a)} \cdot\left(\displaystyle \frac{x^{c}}{x^{a}}\right)^{(c+a-b)} \cdot\left(\displaystyle \frac{x^{a}}{x^{b}}\right)^{(a+b-c)}=$ A. $x^{abc}$ B. $x^{ab + bc + ac}$ C. $\dfrac{1}{x^{abc}}$ D. $1$ Explanation $\begin{aligned} &\left(\frac{x^{b}}{x^{c}}\right)^{(b+c-a)} \cdot\left(\frac{x^{c}}{x^{a}}\right)^{(c+a-b)} \cdot\left(\frac{x^{a}}{x^{b}}\right)^{(a+b-c)} \\\\ =&\left(x^{b-c}\right)^{(b+c-a)}\left(x^{c-a}\right)^{(c+a-b)}\left(x^{a-b}\right)^{(a+b-c)} \\\\ =& x^{b^{2}+b c-a b-b c-c^{2}+a c} \cdot x^{c^{2}+a c-bc-ac-a^{2}+ab} \cdot x^{a^{2}+ab-ac-ab-b^{2}+bc} \\\\ =& x^{b^{2}-c^{2}-a b+ac}\cdot x^{c^{2}-a^{2}-bc+ab}\cdot x^{a^{2}-b^{2}-ac+bc} \\\\ =& x^{0}\\\\ =&1 \end{aligned} $ 5. The simplified form of $ \dfrac{\dfrac{3}{2+\sqrt{3}}-\dfrac{2}{2-\sqrt{3}}}{2-5 \sqrt{3}}$ is A. $1$ B. $0$ C. $2-5\sqrt{3}$ D. $\dfrac{1}{2}-5\sqrt{3}$ Explanation $\begin{aligned} &\dfrac{\dfrac{3}{2+\sqrt{3}}-\dfrac{2}{2-\sqrt{3}}}{2-5 \sqrt{3}}\\\\ &\dfrac{\dfrac{3(2-\sqrt{3})-2(2+\sqrt{3})}{4-3}}{2-5 \sqrt{3}}\\\\ &=\dfrac{6-3 \sqrt{3}-4-2 \sqrt{3}}{2-5 \sqrt{3}}\\\\ &=\dfrac{2-5 \sqrt{3}}{2-5 \sqrt{3}}\\\\ &=1 \end{aligned}$ 6. If $x=5+2 \sqrt{6},$ then $\sqrt{x}-\displaystyle \frac{1}{\sqrt{x}}=$ A. $3\sqrt{2}$ B. $2\sqrt{2}$ C. $\sqrt{3}+\sqrt{2}$ D. $2\sqrt{3}$ Explanation $\begin{aligned} x &=5+2 \sqrt{6} \\\\ \sqrt{x} &=\sqrt{5+2 \sqrt{6}} \\\\ &=\sqrt{2+2 \sqrt{2} \sqrt{3}+3} \\\\ &=\sqrt{(\sqrt{2})^{2}+2 \sqrt{2} \sqrt{3}+(\sqrt{3})^{2}} \\\\ &=\sqrt{(\sqrt{2}+\sqrt{3})^{2}} \\\\ &=\sqrt{2}+\sqrt{3}\\\\ & \sqrt{x}-\dfrac{1}{\sqrt{x}} \\\\ =& \sqrt{2}+\sqrt{3}-\dfrac{1}{\sqrt{2}+\sqrt{3}} \\\\ =& \sqrt{2}+\sqrt{3}-\dfrac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} \\\\ =& \sqrt{2}+\sqrt{3}-\dfrac{\sqrt{2}-\sqrt{3}}{2-3} \\\\ =& \sqrt{2}+\sqrt{3}+\sqrt{2}-\sqrt{3} \\\\ =& 2 \sqrt{2} \end{aligned}$ 7. $\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}=$ A. $16+\sqrt{3}$ B. $16-\sqrt{3}$ C. $2-\sqrt{3}$ D. $2+\sqrt{3}$ Explanation $\begin{aligned} \frac{2+\sqrt{3}}{2-\sqrt{3}} &=\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2 \sqrt{2})(2+\sqrt{3})} \\\\ &=\frac{4+4 \sqrt{3}+3}{4-3} \\\\ &=7+4 \sqrt{3} \\\\ \frac{2-\sqrt{3}}{2+\sqrt{3}} &=\frac{(2-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \\\\ &=\frac{4-4 \sqrt{3}+3}{4-3} \\\\ &=7-4 \sqrt{3}\\\\ \frac{\sqrt{3}-1}{\sqrt{3}+1} &=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \\\\ &=\frac{3-2 \sqrt{3}+1}{3-1} \\\\ \end{aligned}$ $\begin{aligned} & \quad\frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1} \\\\ &=7+4 \sqrt{3}+7-4 \sqrt{3}+2-\sqrt{3} \\\\ &=16-\sqrt{3} \end{aligned}$ 8. If $3^{10} \times 27^{2}=9^{2} \times 3^{n}$ then the value of $n$ is A. $10$ B. $15$ C. $12$ D. $30$ Explanation Solution 8 9. $\sqrt{8}+2 \sqrt{32}-3 \sqrt{128}+4 \sqrt{50} =$ A. $\sqrt{2}$ B. $3\sqrt{2}$ C. $2-3\sqrt{2}$ D. $6 \sqrt{2}$ Explanation $\begin{aligned} & \sqrt{8}+2 \sqrt{32}-3 \sqrt{128}+4 \sqrt{50} \\\\ =& \sqrt{4 \times 2}+2 \sqrt{16 \times 2}-3 \sqrt{64 \times 2}+4 \sqrt{25 \times 2} \\\\ =& 2 \sqrt{2}+8 \sqrt{2}-24 \sqrt{2}+20 \sqrt{2} \\\\ =& 6 \sqrt{2} \end{aligned}$ 10. The value of $(256)^{\frac{5}{4}}$ is A. $64$ B. $512$ C. $2048$ D. $1024$ Explanation $\begin{aligned} &(256)^{5 / 4} \\\\ =&\left(4^{5}\right)^{5 / 4} \\\\ =& 4^{5} \\\\ =& 1024 \end{aligned}$ 11. $(27)^{-\frac{2}{3}}$ lies between A. $1\ \text{and}\ 2$ B. $2\ \text{and}\ 3$ C. $0\ \text{and}\ 1$ D. $0\ \text{and}\ -1$ Explanation $\begin{aligned} &(27)^{-\frac{2}{3}} \\\\ =&\left(3^{3}\right)^{-\frac{2}{3}} \\\\ =& 3^{-2} \\\\ =& \frac{1}{3^{2}} \\\\ =& \frac{1}{9} \\\\ \therefore\ & 0<\frac{1}{9} <1 \end{aligned}$ 12. $\displaystyle \frac{3^{0}+3^{-1}}{3^{-1}-3^{0}}=$ A. $-1$ B. $-2$ C. $2$ D. $0$ Explanation $\begin{aligned} & \dfrac{3^{0}+3^{-1}}{3^{-1}-3^{0}} \\\\ =& \dfrac{7+\dfrac{1}{3}}{\dfrac{1}{3}-7} \\\\ =& \dfrac{\dfrac{4}{3}}{-\dfrac{2}{3}} \\\\ =&-2 \end{aligned}$ 13. $\displaystyle \frac{2}{\sqrt{7}+\sqrt{5}}+\frac{7}{\sqrt{12}-\sqrt{5}}-\frac{5}{\sqrt{12}-\sqrt{7}}=$ A. $0$ B. $1$ C. $-1$ D. $2$ Explanation $\begin{aligned} \dfrac{2}{\sqrt{7}+\sqrt{5}} &=\dfrac{2}{\sqrt{7}+\sqrt{5}} \times \dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}} \\\\ &=\sqrt{7}-\sqrt{5} \\\\ \dfrac{7}{\sqrt{12}-\sqrt{5}} &=\dfrac{7}{\sqrt{12}-\sqrt{5}} \times \dfrac{\sqrt{12}+\sqrt{5}}{\sqrt{12}+\sqrt{5}} \\\\ &=\sqrt{12}+\sqrt{5} \\\\ \dfrac{5}{\sqrt{12}-\sqrt{7}} &=\dfrac{5}{\sqrt{12}-\sqrt{7}} \times \dfrac{\sqrt{12}+\sqrt{7}}{\sqrt{12}+\sqrt{7}} \\\\ &=\sqrt{12}+\sqrt{7}\\\\ \therefore \quad & \dfrac{2}{\sqrt{7}+\sqrt{5}}+\dfrac{7}{\sqrt{12}-\sqrt{5}}-\dfrac{5}{\sqrt{12}-\sqrt{7}} \\\\ =& \sqrt{7}-\sqrt{5}+\sqrt{12}+\sqrt{5}-\sqrt{12}-\sqrt{7} \\\\ =& 0 \end{aligned}$ 14. $\sqrt[3]{2^{4} \sqrt{2^{-5} \sqrt{2^{6}}}}=$ A. $2^{\frac{5}{3}}$ B. $2^5$ C. $2$ D. $1$ Explanation $\begin{aligned} &\ \sqrt[3]{2^{4} \sqrt{2^{-5} \sqrt{2^{6}}}} \\\\ =&\ \sqrt[3]{2^{4} \sqrt{2^{-5}\left(2^{3}\right)}} \\\\ =&\ \sqrt[3]{2^{4} \sqrt{2^{-2}}} \\\\ =&\ \sqrt[3]{2^{4}\left(2^{-1}\right)} \\\\ =&\ \sqrt[3]{2^{3}}\\\\ =&\ 2 \end{aligned}$ 15. The value of $5^{\frac{1}{4}} \times(125)^{0.25}$ is A. $5$ B. $5\sqrt{5}$ C. $\sqrt{5}$ D. $25$ Explanation $\begin{aligned} &\ 5^{\frac{1}{4}} \times(125)^{0.25} \\\\ =&\ 5^{\frac{1}{4}} \times\left(5^{3}\right)^{\frac{1}{4}} \\\\ =&\ 5^{\frac{1}{4}} \times 5^{\frac{3}{4}} \\\\ =&\ 5^{\frac{1}{4}+\frac{3}{4}} \\\\ =&\ 5 \end{aligned}$ 16. If $10^{x}=\displaystyle \frac{1}{2}$ then $10^{-8 x}=$ A. $\dfrac{1}{256}$ B. $256$ C. $64$ D. $\dfrac{1}{64}$ Explanation $\begin{aligned} 10^{x} &=\frac{1}{2} \\\\ 10^{-8 x} &=\left(10^{x}\right)^{-8} \\\\ &=\left(\frac{1}{2}\right)^{-8} \\\\ &=2^{8} \\\\ &=256 \end{aligned}$ 17. Given that $\sqrt{5}=2.236$ and $\sqrt{3}=1.732,$ then the value of $\displaystyle \frac{1}{\sqrt{5}+\sqrt{3}}$ is A. $0.564$ B. 0.504$ C. $0.202$ D. $0.252$ Explanation $\begin{aligned} \sqrt{5} &=2.236 \\\\ \sqrt{3} &=1.732 \\\\ \frac{1}{\sqrt{5}+\sqrt{3}} &=\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \\\\ &=\frac{\sqrt{5}-\sqrt{3}}{2} \\\\ &=\frac{1}{2}(0.504) \\\\ &=0.252 \end{aligned}$ 18. If $2^{x}=\sqrt[3]{32}$, then $x=$ A. $\dfrac{5}{3}$ B. $\dfrac{3}{5}$ C. $5$ D. $3$ Explanation $\begin{aligned} &2^{x}=\sqrt[3]{32} \\\\ &2^{x}=(2)^{\frac{5}{3}} \\\\ &x=\frac{5}{3} \end{aligned}$ 19. $\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=$ A. $2\sqrt{6}$ B. $4\sqrt{6}$ C. $3\sqrt{2}$ D. $2\sqrt{3}$ Explanation $\begin{aligned} &\ \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\\\ =&\ \frac{(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} \\\\ =&\ \frac{3+2 \sqrt{6}+2-3+2 \sqrt{6}-2}{3-1} \\\\ =&\ 4 \sqrt{6} \end{aligned}$ 20. If $3^{(x+y)}=81$ and $81^{(x-y)}=3,$ then the value of $x$ is A. $\dfrac{1}{8}$ B. $\dfrac{15}{8}$ C. $\dfrac{17}{8}$ D. $42$ Explanation $\begin{aligned} 3^{x+y} &=81 \\\\ 3^{x+y} &=3^{4} \\\\ x+y &=4 \ldots(1) \\\\ 81^{x-y} &=3 \\\\ \left(3^{4}\right)^{x-y} &=3 \\\\ 4(x-y) &=3 \\\\ 3^{4(x-y)} &=1 \\\\ x-y &=\frac{1}{4} \ldots(2)\\\\ (1)+(2), \\\\ 2 x=\frac{17}{4} \\\\ x=\frac{17}{8} \end{aligned}$ Submit answers Your Score: စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!
