Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(ĐK:x\ne\pm1;x\ne0;x\ne3\)
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
b) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}\)
\(=4x-\sqrt{8}+\frac{x\left(x+2\right)}{x+2}\)
\(=4x-\sqrt{8}+x\)
\(=5x-\sqrt{8}\)
Với \(x=-\sqrt{2}\) ta có:
\(5x-\sqrt{8}=5\cdot\left(-\sqrt{2}\right)-\sqrt{4\cdot2}=-5\sqrt{2}-2\sqrt{2}=-7\sqrt{2}\)
a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:
\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)
\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)
b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)
=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)
c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)
d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6
a)
2x-4=2(x-2)
2x+4=2(x+2)
x
Để P xác định thì
[2(x-2) => [2(x+2)
[2(x+2) =>[ 2(x-2)
[ (x-2)(x+2) => [(x+2)(x-2)
Vay 2(x+2) , 2(x-2), (x+2)(x-2) thi P xác định
a) \(H=\left(\frac{x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{x^2-2x+4}{x^2-4}\right).\frac{x+3}{x+2}\)
\(=\left(\frac{x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{x^2-2x+4}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+3}{x+2}\)
\(=\left(\frac{x^2+2x}{\left(x+2\right)^2}-\frac{\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right).\frac{x+3}{x+2}\)
\(=\frac{-4}{\left(x+2\right)^2}.\frac{x+3}{x+2}=\frac{-4x-12}{\left(x+2\right)^3}\)