\(a,\left[\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2\right]:\left[2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}\right]\)

\(=\left[\left(-\frac{1}{8}\right)-\frac{27}{64}.4\right]:\left[2.\left(-1\right)+\frac{9}{16}-\frac{3}{8}\right]\)

\(=\left[\left(-\frac{1}{8}-\frac{27}{16}\right)\right]:\left[-2+\frac{9}{16}-\frac{3}{8}\right]\)

\(=\frac{-2-27}{16}:\frac{-32+9-6}{16}\)

\(=-\frac{29}{16}:\frac{-29}{16}=1\)

\(b,\left[\left(\frac{4}{3}\right)^{-2}\left(\frac{3}{2}\right)^4\right]:\left(\frac{3}{2}\right)^6\)

\(=\left(\frac{9}{16}.\frac{81}{16}\right):\frac{729}{64}\)

\(=\frac{729}{64}:\frac{729}{64}=1\)

\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)

\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow27x-2x-4x-27+2=0\)

\(\Leftrightarrow21x=25\)

\(\Leftrightarrow x=\frac{25}{21}\)

Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !

\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)

\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)

\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)

\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)

\(\Leftrightarrow-20x-12=56\)

\(\Leftrightarrow-20x=68\)

\(\Leftrightarrow x=-\frac{17}{5}\)

Tự check lại nhá

\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)

\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)

\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

Ta dễ thấy \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}< 0\) nên

\(x+95=0\Leftrightarrow x=-95\)

a) ĐKXĐ : \(x\ne0\)

\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=\frac{-5}{4}\)

\(\left(\frac{-9x}{3x}+\frac{9}{3x}-\frac{x}{3x}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=\frac{-5}{4}\)

\(\frac{-9x+9-x}{3x}:\frac{15+6+10}{15}=\frac{-5}{4}\)

\(\frac{-10x+9}{3x}:\frac{31}{15}=\frac{-5}{4}\)

\(\frac{-10x+9}{3x}=\frac{-31}{12}\)

\(\Leftrightarrow12\left(-10x+9\right)=-31\cdot3x\)

\(\Leftrightarrow-120x+108=-93x\)

\(\Leftrightarrow-120x+93x=-108\)

\(\Leftrightarrow-27x=-108\)

\(\Leftrightarrow x=4\)

b) ĐKXĐ : \(x\ne0\)

\(\frac{-3x}{4}\cdot\left(\frac{1}{x}+\frac{2}{7}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{-3x}{4}=0\\\frac{1}{x}+\frac{2}{7}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\\frac{-2}{-2x}=\frac{-2}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\x=\frac{-7}{2}\end{cases}}\)

Vậy.....

c) phân tích ra rồi làm thôi e :)) a bận rồi 

Trần Thanh Phương làm đúng rồi đó !

k bạn ý đi !

a) ĐKXĐ: x khác +2

\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)

<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)

<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22

<=> x^2 - 7x - 2 = 2x - 22

<=> x^2 - 7x - 2 - 2x + 22 = 0

<=> x^2 - 9x + 20 = 0

<=> (x - 4)(x - 5) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 hoặc x = 5

làm nốt đi 

Giải:

1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)

\(=7^8.\left(\dfrac{1}{7}\right)^8\)

\(=7^8.\dfrac{1^8}{7^8}\)

\(=1\)

2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)

\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)

\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)

\(=1\)

3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)

\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)

\(=1\)

4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)

\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)

\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)

\(=\dfrac{5}{13}\)

Vậy ...

Bài 1:

a) \(-\frac{4}{5}-\frac{8}{25}\left(\frac{-5}{2}-0,125\right)\\ =-\frac{4}{5}-\frac{8}{25}\left(\frac{-5}{2}-\frac{1}{8}\right)\\ =-\frac{4}{5}-\frac{8}{25}\left(\frac{-20}{8}-\frac{1}{8}\right)\\ =-\frac{4}{5}-\frac{8}{25}\cdot\frac{-21}{8}\\ =-\frac{4}{5}-\frac{-21}{25}\\ =\frac{-4}{5}+\frac{21}{25}\\ =\frac{-20}{25}+\frac{21}{25}=\frac{1}{25}\)

c) \(5\frac{1}{2}-4\frac{2}{3}:\frac{16}{9}-3\frac{1}{3}:\frac{16}{9}\\ =5\frac{1}{2}-\left(4\frac{2}{3}:\frac{16}{9}+3\frac{1}{3}:\frac{16}{9}\right)\\ =5\frac{1}{2}-\left(4\frac{2}{3}+3\frac{1}{3}\right):\frac{16}{9}\\ =5\frac{1}{2}-8\cdot\frac{9}{16}\\ =\frac{11}{2}-\frac{9}{2}=\frac{2}{2}=1\)

Bài 2:

a) \(\left(20\%x+\frac{2}{5}x-2\right):\frac{1}{3}=-2013\\ \left(\frac{1}{5}x+\frac{2}{5}x-2\right)\cdot3=-2013\\ \left[x\left(\frac{1}{5}+\frac{2}{5}\right)-2\right]=\left(-2013\right):3\\ x\cdot\frac{3}{5}-2=-671\\ x\cdot\frac{3}{5}=-671+2\\ x\cdot\frac{3}{5}=-669\\ x=\left(-669\right):\frac{3}{5}\\ x=\left(-669\right)\cdot\frac{5}{3}\\ x=-1115\)Vậy x = -1115

b) \(\left(4,5-2\left|x\right|\right)\cdot1\frac{4}{7}=\frac{11}{14}\\ \left(\frac{9}{2}-2\left|x\right|\right)\cdot\frac{11}{7}=\frac{11}{14}\\ \frac{9}{2}-2\left|x\right|=\frac{11}{14}:\frac{11}{7}\\ \frac{9}{2}-2\left|x\right|=\frac{11}{14}\cdot\frac{7}{11}\\ \frac{9}{2}-2\left|x\right|=\frac{1}{2}\\ 2\left|x\right|=\frac{9}{2}-\frac{1}{2}\\ 2\left|x\right|=4\\ \left|x\right|=4:2\\ \left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)Vậy x ∈ {2 ; -2}