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![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ne\pm1;x\ne0;x\ne3\)
Với \(x\ne\pm1;x\ne0;x\ne3\)thì\(M=\frac{x^3+2x^2-x-2}{x^3-2x^2-3x}\left[\frac{\left(x+2\right)^2-x^2}{4x^2-4}-\frac{3}{x^2-x}\right]=\frac{x^2\left(x+2\right)-\left(x+2\right)}{\left(x^3-x\right)-\left(2x^2+2x\right)}\left[\frac{x^2+4x+4-x^2}{4x^2-4}-\frac{3}{x\left(x-1\right)}\right]\)\(=\frac{\left(x-1\right)\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)\left(x-1\right)-2x\left(x+1\right)}\left[\frac{4\left(x+1\right)}{4\left(x+1\right)\left(x-1\right)}-\frac{3}{x\left(x-1\right)}\right]=\frac{\left(x-1\right)\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-3x\right)}\left[\frac{1}{x-1}-\frac{3}{x\left(x-1\right)}\right]\)\(=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x-3\right)}.\frac{x-3}{x\left(x-1\right)}=\frac{x+2}{x^2}\)
M = 3 \(\Leftrightarrow\frac{x+2}{x^2}=3\Leftrightarrow3x^2-x-2=0\Leftrightarrow\left(x-1\right)\left(3x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}\)
Mà \(x\ne1\)(theo điều kiện) nên x =-2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{x^3-1}{x^2-x}+\frac{x^2-4}{x^2-2x}-\frac{2-x}{x}\right)\div\frac{x+1}{x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x^2+x+1}{x}+\frac{x+2}{x}-\frac{2-x}{x}\right)\times\frac{x}{x+1}\)
\(=\left(\frac{x^2+x+1+x+2-2+x}{x}\right)\times\frac{x}{x+1}\)
\(=\frac{x^2+3x+1}{x}\times\frac{x}{x+1}=\frac{x^2+3x+1}{x+1}\)
b) x3 - 4x2 + 3x = 0
<=> x( x2 - 4x + 3 ) = 0
<=> x( x - 1 )( x - 3 ) = 0
<=> x = 0 (ktm) hoặc x = 1(tm) hoặc x = 3(tm)
Bạn tự thế các giá trị tm nhé ;)
b) Ta có: \(x^3-4x^2+3x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
<=> x=0 ( loại) hoặc x=1 (loại) hoặc x=3 ( thỏa mãn)
Thay x=3 vào A ta có:
\(A=\frac{3^2+3.3+1}{3+1}=\frac{19}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
a) \(đkxd:x\ne2;x\ne-2;x\ne0;x\ne3\)
Ta có: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\left(\frac{\left(x+2\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right]:\frac{x-3}{x\left(2-x\right)}\)
\(A=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x^2}{x-3}\)
b) Ta có: \(4x^2>0\left(\forall x\ne0\right)\)
=> Để A>0 thì \(x-3>0\)
\(\Rightarrow x>3\)
Vậy với \(x>3\)thì A>0
c) Ta có: \(\left|x-7\right|=4\)\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=11\\x=3\end{cases}}\)
Mà theo điều kiện xác định, \(x\ne3\)
\(\Rightarrow x=11\)
Khi đó, \(A=\frac{4.11^2}{11-3}=\frac{121}{2}\)
Vậy \(A=\frac{121}{2}\)
Học tốt!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
đkxd: \(x\ne\left\{\pm3\right\}\)
a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)
=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)
=\(\frac{x+12}{x-3}\)
b)|2x+1|=5
<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2
với x=-3 thì B=\(\frac{-3}{2}\)
với x=2 thì B=-14