1.(25¹⁵.4¹⁵)/(5¹⁷.20¹⁶)=(5³⁰.2³⁰)/(5¹⁷.5¹⁶.2³²)=1/(5³.2²)=1/500

2.a,-x/2=8/-x=>-x.(-x)=2*8 =>x^2=16=(-4)^2=4^2

=>x=4 hoặc x=-4

b,(3/4)^2x/(2/5)¹⁰=(15/8)¹⁰

(3/4)^2x=(2/5*15/8)¹⁰

(3/4)^2x=(3/4)¹⁰

2x=10

x=5

1)  \(\frac{25^{15}\cdot4^{15}}{5^{17}\cdot20^{16}}\)

\(=\frac{5^{30}\cdot2^{30}}{5^{33}\cdot2^{32}}\)

\(=\frac{1}{5^3\cdot2^2}\)

\(=\frac{1}{500}\)

2) 

a) \(\frac{-x}{2}=\frac{8}{-x}\)

\(\Rightarrow\left(-x\right)\left(-x\right)=8\cdot2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=\left\{\pm4\right\}\)

2b) \(\left(\frac{3}{4}\right)^{2x}:\left(\frac{2}{5}\right)^{10}=\left(\frac{15}{8}\right)^{10}\)

\(\left(\frac{3}{4}\right)^{2x}=\left(\frac{15}{8}\right)^{10}\cdot\left(\frac{2}{5}\right)^{10}\)

\(\left(\frac{3}{4}\right)^{2x}=\left(\frac{15}{8}\cdot\frac{2}{5}\right)^{10}=\left(\frac{3}{4}\right)^{10}\)

\(\left(\frac{3}{4}\right)^{2x}=\left(\frac{3}{4}\right)^{2\cdot5}\)

\(\Rightarrow x=5\)

A=3/4

B=29/35

mình nghĩ thế

Sai thôi nhé!

\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)

\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)

\(A=1+\frac{1}{2}\)

\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)

\(A=\frac{3}{2}\)

Bạn viết đề thiếu trầm trọng quá !!!

Đáp án: thiếu đề

@#@

mời bn xem xét lại đề bài.

~hok tốt~

Bài 1:
a)

\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{(2^3)^{20}+(2^2)^{20}}{(2^2)^{25}+(2^6)^{5}}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}(2^{20}+1)}{2^{30}(2^{20}+1)}=2^{10}\)

b)

\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{(3^2.5)^{10}.5^{20}}{(3.5^2)^{15}}=\frac{3^{20}5^{30}}{3^{15}.5^{30}}=\frac{3^{20}}{3^{15}}=3^5\)

Bài 2:

Ta thấy $(x-2)^{2012}=[(x-2)^{1006}]^2\geq 0$ với mọi $x\in\mathbb{R}$

$|b^2-9|^{2014|\geq 0$ với mọi $b\in\mathbb{R}$ (tính chất trị tuyệt đối)

Do đó để tổng của chúng bằng $0$ thì:

\((x-2)^{2012}=|b^2-9|^{2014}=0\)

\(\Leftrightarrow \left\{\begin{matrix} x-2=0\\ b^2-9=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ b=\pm 3\end{matrix}\right.\)

Vậy.......