Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đề sai ạ ! Sửa nhé :
\(S=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^3-4x}{2x^2-x^3}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)}{x-2}+\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(x^2-4\right)}{x^2\left(2-x\right)}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)^2+4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)}{-x\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x+2\right)\left(x-2\right)}.\frac{-x}{\left(x+2\right)}\)
\(\Leftrightarrow S=\frac{-x\left(4x^2-8x\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2}{\left(x+2\right)^2}\)
P/s : nếu làm theo đề của bạn, sẽ ra kq dài... Nên mik tiện sửa, còn nếu đề bạn đúng rồi thì mik sẽ làm lại ạ !
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\) ĐKXD: \(x\ne\pm2,x\ne0,x\ne3\)
\(\Leftrightarrow\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{2-x}{2+x}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(\Leftrightarrow\left(\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right):\left(\frac{x-3}{x\left(2-x\right)}\right)\)
\(\Leftrightarrow\left(\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\right)\cdot\left(\frac{x\left(2-x\right)}{x-3}\right)\)
\(\Leftrightarrow\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2-x\right)}{x-3}\)
\(\Leftrightarrow\frac{4x^2}{x-3}\)
b, Để A>0 thì \(\frac{4x^2}{x-3}>0\)
\(\Rightarrow4x^2>0\)
\(\Rightarrow x>0\)
c, Ta có
\(\left|x-7\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\left(l\right)\end{cases}}}\)
Với \(x=11\Rightarrow\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)
\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)
\(A=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-3x}\right)\)
\(=\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x^2-3x}{x^2-3x}\)
\(=\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2x-3\right)}{x\left(x-3\right)}\)
\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)
\(=\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}\)
b
Xét hơi bị nhiều TH nhá:(
Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)
TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)
\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)
Bạn tự xét nốt nhá!
c
\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)
\(\Rightarrow x=11;x=3\)
Thay vào .....
