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\(\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}+\frac{1\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(=\frac{x+\left(x-2\right)-2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\div\left(1-\frac{x}{x+2}\right)\)
\(=\frac{x-2}{1}\div\left(1-\frac{x}{x+2}\right)\)
\(=\frac{x-2}{1}\div\left(\frac{x+2}{x+2}-\frac{x}{x+2}\right)\)
\(=\frac{x-2}{1}\div\left(\frac{x+2-x}{x+2}\right)=\frac{x-2}{1}\div\frac{2}{x+2}\)
\(=\frac{x-2}{1}\times\frac{x+2}{2}=\frac{\left(x-2\right)\left(x+2\right)}{1.2}=\frac{x^2-2^2}{2}=\frac{x^2-2}{1}=x^2-2\)
(Sai thì thôi)
#Học tốt!!!
~NTTH~
ĐKXĐ: \(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
\(P=\left(\frac{x+1}{x-2}-\frac{2x}{x+2}-\frac{x^2-x}{4-x^2}\right):\left(3-\frac{3x+4}{x+2}\right)\)
\(=\left[\frac{x+1}{x-2}-\frac{2x}{x+2}+\frac{x^2-x}{\left(x+2\right)\left(x-2\right)}\right]:\left(3-\frac{3x+4}{x+2}\right)\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)-2x\left(x-2\right)+x^2-x}{\left(x+2\right)\left(x-2\right)}\right]:\left(\frac{3x+6-3x-4}{x+2}\right)\)
\(=\left(\frac{x^2+3x+2-2x^2+4x+x^2-x}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{2}\)
\(=\frac{6x+2}{x-2}.\frac{1}{2}=\frac{3x+1}{x-2}\)
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)
b) Với \(M=\frac{1}{4}\)ta có :
\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)
\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}=\frac{1}{x-2}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)
=\(\left(1+\frac{4}{x-2}\right):\left(\frac{x^2-4}{2}\right)\)
=\(\left(\frac{x-2}{x-2}+\frac{4}{x-2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{2}\right)\)
=\(\frac{x-2+4}{x-2}\cdot\frac{2}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{x+2}{x-2}.\frac{2}{\left(x-2\right).\left(x+2\right)}\)
=\(\frac{2}{\left(x-2\right)^2}\)
Mình quên ĐKXĐ
Sorry bạn nha