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\(\text{(x+1)^3 -(x-1)^3-(x^3-1)-(x-1)(x^2+x+1)}\)
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10 tháng 11 2021
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
10 tháng 11 2021
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
18 tháng 12 2021
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
19 tháng 1 2022
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
6 tháng 9 2017
\(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x+x^2}{1-x}.\frac{\left(1-x\right)-x^2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x\left(1-x\right)}{1-x}.\frac{\left(1-x\right)\left(1-x^2\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{\left(1-x\right)\left(1+x^2\right)}{1-x}.\frac{\left(1-x\right)\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\left(1+x^2\right)\left(1-x\right)\)
\(=-x^3+x^2-x+1\)
6 tháng 9 2017
Ta có : \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{\left(1-x\right)}-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-\left(x^2-x^3\right)}\)
\(=\left(\left(1+x+x^2\right)-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-x^2\left(x-1\right)}\)
\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x^2\right)}\)
\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x\right)\left(x+1\right)}\)
\(=\left(1+x^2\right):\frac{1}{1-x}\)
\(=\left(1+x^2\right)\left(1-x\right)\)
10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
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\(\left(x+1\right)^3-\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)\\ =\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-\left(x^3-1\right)\\ =2.\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-\left(x^3-1\right)\\ =2.\left(3x^2+1\right)-\left(x^3-1\right)\\ =6x^2+2-x^3+1=-x^3+6x^2+3\)
Ghi thiếu (x^3-1) kìa bạn