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a: ĐKXĐ: x<>1; x<>-1
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1}{x-1}-\dfrac{1}{x-1}=\dfrac{x}{x-1}\)
b: x^2+3x+2=0
=>x=-1(loại) hoặc x=-2(nhận)
Khi x=-2 thì A=-2/(-3)=2/3
\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)
\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)
\(A=\left(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}\right).\dfrac{2x+6}{8x}\)
\(a,\) Điều kiện xác định: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\8x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\\x\ne0\end{matrix}\right.\)
\(b,A=\left(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}\right).\dfrac{2x+6}{8x}\)
\(=\left[\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{2\left(x+3\right)}{8x}\)
\(=\dfrac{\left(x-3-x-3\right)\left(x-3+x+3\right)}{\left(x+3\right)\left(x-3\right)}.\dfrac{x+3}{4x}\)
\(=\dfrac{-6.2x}{\left(x-3\right)}.\dfrac{1}{4x}\)
\(=\dfrac{-12x}{4x\left(x-3\right)}\)
\(=\dfrac{-3}{x-3}\)
\(c,A=\dfrac{1}{2}\Rightarrow\dfrac{-3}{x-3}=\dfrac{1}{2}\Leftrightarrow x=-3\)
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
Bài 1:
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)
Bài 2:
\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)
a: \(M=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right):\dfrac{x-1-x+3}{x-1}\)
\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2}\)
\(=\dfrac{-2x+2}{2\left(x+1\right)}\cdot\dfrac{1}{2}=\dfrac{-x+1}{2}\)
b: Thay x=-1/2 vào M, ta được:
\(M=\dfrac{\dfrac{1}{2}+1}{2}=\dfrac{3}{2}:2=\dfrac{3}{4}\)
a, \(M=\left(\dfrac{x^2-1-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-1-x+3}{x-1}\right)\)
\(=\left(\dfrac{-1+x-3x-3}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{2}{x-1}=\dfrac{-2x-4}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-\left(x+2\right)}{2\left(x+1\right)}\)
b, Thay x =-1/2 vào ta đc
\(-\dfrac{\left(\dfrac{-1}{2}+2\right)}{2\left(-\dfrac{1}{2}+1\right)}=\dfrac{-\dfrac{3}{2}}{2\left(\dfrac{1}{2}\right)}=\dfrac{-3}{2}\)