((x+x^3/x^2+1)-(x-x^3/x^2-1))/((1+x/1-x)-(1-x/1+x)) với x khác 1, x khác-1 và x khác 0
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MN
18 tháng 4 2019
Ta có: \(2\left(x-\frac{1}{2}\right)+3\left(-1+\frac{x}{3}\right)=x\left(\frac{2}{x}-1\right)\)
\(2x-1+-3+\frac{3x}{3}=\frac{2x}{x}-x\)
\(2x-1+-3+x=2-x\)
\(\left(2x+x\right)+\left(-3\right)-1=2-x\)
\(3x+\left(-4\right)=2-x\)
\(3x+x=2-\left(-4\right)\)
\(4x=6\)
\(x=6:4\)
\(x=\frac{6}{4}=\frac{3}{2}\)
A
1
20 tháng 2 2022
\(P=\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3-x^2+x-1}\right):\left(\dfrac{1-2x}{x+1}\right)\left(ĐKXĐ:x\ne0;x\ne\pm1\right)\)
\(=\left(\dfrac{1}{x-1}-\dfrac{2x}{x^2\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1-2x}{x+1}\right)\)
\(=\left(\dfrac{1}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x^2+1\right)}\right):\left(\dfrac{1-2x}{x+1}\right)\)
\(=\left(\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\right):\left(\dfrac{1-2x}{x+1}\right)\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}:\dfrac{1-2x}{x+1}\)
\(=\dfrac{x-1}{x^2+1}:\dfrac{1-2x}{x+1}\)
\(=\dfrac{x-1}{x^2+1}.\dfrac{x+1}{1-2x}\)
\(=\dfrac{x^2-1}{\left(x^2+1\right)\left(1-2x\right)}\)
2 tháng 10 2021
\(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\left(x>0;x\ne1;x\ne\dfrac{1}{4}\right)\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
2 tháng 10 2021
a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)