giúp e vs ạ e cần gấp
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23 tháng 12 2021
g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
23 tháng 12 2021
\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
Bài I
\(1,=x\left(x-y\right)\\ 2,=x\left(y+1\right)+y+1=\left(x+1\right)\left(y+1\right)\\ 3,=x\left(x^2-2x-5x+10\right)=x\left(x-2\right)\left(x-5\right)\)
Bài II
\(1,=x-x^2+x^2-x-2=-2\\ 2,\Leftrightarrow x^2+6x+9-x^2=45\\ \Leftrightarrow6x=36\Leftrightarrow x=6\)
Bài III
\(1,A=\dfrac{4-9}{3\left(2+5\right)}=\dfrac{-5}{3\cdot7}=-\dfrac{5}{21}\\ 2,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\\ 3,P=AB=\dfrac{\left(x-3\right)\left(x+3\right)}{3\left(x+5\right)}\cdot\dfrac{3}{x+3}=\dfrac{x-3}{x+5}=1-\dfrac{8}{x+5}\in Z\\ \Leftrightarrow x+5\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow x\in\left\{-13;-9;-7;-6;-4;-1;3\right\}\)