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\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow A=\left(\frac{-1}{x-1}+\frac{2}{x+1}+\frac{5-x}{x^2-1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\left[\frac{-x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2x-2}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\frac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2}\)
\(\Leftrightarrow A=\frac{2\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=1\)
vậy \(A=1\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}:\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}\cdot\frac{^2-1}{1-2x}=\frac{2}{1-2x}\)ĐK: x khác 1/2
Answer:
a) \(Q=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{4-2x}{x^3-x^2+x}\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right).\frac{x\left(x^2-x+1\right)}{4-2x}\)
\(=\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x\left(x^2-x+1\right)}{2\left(2-x\right)}\)
\(=\frac{\left(-2x^2+4x\right)-x}{\left(x+1\right)-2\left(2-x\right)}\)
\(=\frac{+2x^2\left(-x+2\right)}{\left(x+1\right)-2\left(2-x\right)}\)
\(=\frac{x^2}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=\frac{-5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}Q=\frac{4}{3}\\Q=\frac{1}{2}\end{cases}}\)
Bài làm
a) \(Q=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{4-2x}{x^3-x^2+x}\)
\(Q=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\frac{4-2x}{x^3-x^2+x}\)
(bước trên là mình đổi dấu ở phân số thứ hai, dấu âm chuyển xuống dưới mẫu nên đổi dấu ở mẫu, sau đó nhân với cả cụm x + 1 nha, tại hơi tắt nên thêm dòng giải thích cho dễ hiểu)
\(Q=\left(\frac{x+1}{x^3+1}+\frac{x+1}{x^3+1}-\frac{2x^2-2x+2}{x^3+1}\right):\frac{4-2x}{x^3-x^2+x}\)
\(Q=\frac{-2x^2+4x}{x^3+1}\cdot\frac{x\left(x^2-x+1\right)}{4-2x}\)
\(Q=\frac{x\left(4-2x\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{x\left(x^2-x+1\right)}{4-2x}\)
\(Q=\frac{x^2}{x+1}\)
b) Ta có: \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
=> \(x-\frac{3}{4}=\pm\frac{5}{4}\)
=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}}\)
*Trường hợp 1: Khi x = 2
Thay x = 2 vào \(Q=\frac{x^2}{x+1}\)ta được:
\(Q=\frac{2^2}{2+1}=\frac{4}{3}\)
Vậy khi x = 2 thì Q = 4/3
*Trường hợp 2: Khi x = -1/2
Thay x = -1/2 vào \(Q=\frac{x^2}{x+1}\)ta được:
\(Q=\frac{\left(-\frac{1}{2}\right)^2}{-\frac{1}{2}+1}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}:\frac{1}{2}=\frac{1}{4}\cdot2=\frac{1}{2}\)
Vậy x = -1/2 thì Q = 1/2
\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{x^2-1}.\frac{x^2-1}{5}=\)
\(=\frac{2x^2+3x+1+8-x^2+2x-1}{5}=\frac{x^2+5x+8}{5}\)
\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\left(x\ne\pm1\right)\)
\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{2x^2+3x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{2x^2+3x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{\left(x^2+5x+8\right)\cdot\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)5}=\frac{x^2+5x+8}{5}\)
a, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b, Thay x = -2 ta được :
\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)
Vậy A nhận giá trị 1/3
\(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right)\div\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)