đề đủ không đó sao mk thấy khó giải thế như kiểu thiếu dự kiện á

a, P= \(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\): ( \(\frac{x+1}{x}\)+ \(\frac{1}{x-1}\)- \(\frac{x^2-2}{x\left(x-1\right)}\)

P= \(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\): \(\frac{\left(x+1\right)\left(x-1\right)+x-x^2+2}{x\left(x-1\right)}\)

P= \(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\). \(\frac{x\left(x-1\right)}{x^2-1+x-x^2+2}\)

P=  \(\frac{x^2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)

P= \(\frac{x^2}{x-1}\)( đkxđ x khác 1)

b, để P=\(\frac{-1}{2}\)\(\Rightarrow\)\(\frac{x^2}{x-1}\)=\(\frac{-1}{2}\)\(\Rightarrow\)1-x  =  2x\(^2\)

\(\Rightarrow\)2x\(^2\)+ x-1 = 0\(\Rightarrow\)2x\(^2\)- 2x +x - 1   =0\(\Rightarrow\)(x -1 ) (2x + 1) = 0

\(\Rightarrow\)\(\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\orbr{\begin{cases}x=1\left(ktm\right)\\x=\frac{-1}{2}\left(tm\right)\end{cases}}\)

vậy x= \(\frac{-1}{2}\)

c, tớ chịu thôi mà tớ mỏi tay lắm òi. k cho tớ nhé

1  The boy who is smart was bitten by a dog
2  The vase which was my favourite was broken yesterday
 

hello

\(\frac{x+2}{\left(x-2\right)^2}=\frac{16\left(x+2\right)}{16\left(x-2\right)^2}=\frac{16x+32}{16\left(x-2\right)^2}=\frac{x^2+12x+36-x^2+4x-4}{16\left(x-2\right)^2}\)

\(=\frac{\left(x+6\right)^2}{16\left(x-2\right)^2}-\frac{1}{16}\ge-\frac{1}{16}\)

Dấu = xảy ra khi x=-6

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x=2-2\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::