Bài làm
a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\)
\(P=\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(P=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(P=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x+2}\)
\(P=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{x+1}\)
\(P=\frac{x+1}{x-2}\)
b) Thay \(x=\frac{1}{2}\)vào P ta được:
\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}\)
\(P=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{2}{2}}\)
\(P=\frac{3}{2}:\frac{-1}{2}\)
\(P=\frac{3}{2}.\left(-2\right)\)
\(P=-3\)
Vậy giá trị của \(P=-3\) tại \(x=\frac{1}{2}\)
a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)
\(\Leftrightarrow P=\left(\frac{x}{x-2}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(\Leftrightarrow P=\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(\Leftrightarrow P=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(\Leftrightarrow P=\frac{x^2+2x+1}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x+2}{x+1}\)
\(\Leftrightarrow P=\frac{\left(x+1\right)^2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{x+1}{x-2}\)
Vậy \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)
b) Ta có \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)
Thay x=\(\frac{1}{2}\left(tm\right)\)vào P ta có:
\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{4}{2}}=\frac{\frac{3}{2}}{\frac{-3}{2}}=\frac{3}{2}:\frac{-3}{2}=-1\)
Vậy \(P=-1\)khi x=\(\frac{1}{2}\)
xin mn, cái đoạn cuối mik phân tích là 1/2 - 2 ý, là pk bằng 1/2 - 4/2 nha, tức là bn heo jj đó lm đúng :> sai cách quy đồng sương sương, chán ghê :>
