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Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a/ \(B=\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\left(\dfrac{x+y}{x^2+xy+y^2}+\dfrac{1}{x-y}\right)\)
\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{\left(x+y\right)\left(x-y\right)+x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{x^2-y^2+x^2+xy+y^2}{x^3-y^3}\)
\(=\dfrac{2x^2+xy}{xy}=\dfrac{x\left(2x+y\right)}{xy}=\dfrac{2x+y}{y}\)
b/ Khi x = -1/2 và y = 3 ta có:
\(B=\dfrac{2\cdot\left(-\dfrac{1}{2}\right)+3}{3}=\dfrac{-1+3}{3}=\dfrac{2}{3}\)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
a) Rút gọn A
\(A=\left(\dfrac{1}{2-x}+\dfrac{1}{2+x}\right):\left(\dfrac{1}{2-x}-\dfrac{1}{2+x}\right)+\dfrac{2}{2+x}\)
ĐKXĐ : \(\left\{{}\begin{matrix}2-x\ne0\\2+x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.andx\ne0\)
Ta có : \(A=\left(\dfrac{2+x}{\left(2-x\right)\left(2+x\right)}+\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right):\left(\dfrac{2+x}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right)+\dfrac{2}{2+x}\)
\(A=\dfrac{4}{\left(2-x\right)\left(2+x\right)}.\dfrac{\left(2-x\right)\left(2+x\right)}{2+x}+\dfrac{2}{2+x}\)
\(A=\dfrac{4}{2+x}+\dfrac{2}{2+x}\)
\(A=\dfrac{6}{2+x}\)
a) ĐKXĐ: \(x\ne0;x\ne\pm1\)
Ta có: \(A=\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{x}{x+1}\)
\(A=\left[\dfrac{1}{x\left(x+1\right)}-\dfrac{2-x}{x+1}\right].\dfrac{x}{x+1}\)
\(A=\dfrac{1-x\left(2-x\right)}{x\left(x+1\right)}.\dfrac{x}{x+1}\)
\(A=\dfrac{1-2x+x^2}{x\left(x+1\right)}.\dfrac{x}{x+1}\)
\(A=\dfrac{\left(1-x\right)^2}{\left(x+1\right)^2}\)