Rút gọn biểu thức: x + 2 2 x - 4 = x - 2 2 x - 4 v ớ i x ≥ 2
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VN
0
24 tháng 9 2021
\(i,=\left(x-3\right)\left(x+3\right)^2-\left(x-3\right)\left(x^2+3x+9\right)\\ =\left(x-3\right)\left(x^2+6x+9-x^2-3x-9\right)\\ =3x\left(x-3\right)=3x^2-9x\\ ii,=x^3-8-25-x^3=-33\)
24 tháng 9 2021
ii: Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)
\(=x^3-8-x^3-25\)
=-33
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
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
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 \(x\in\left\{3;1;5;-1\right\}\)
N
23 tháng 11 2021
\(\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{4-2\sqrt{x}+4\sqrt{x}+2x-2x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
14 tháng 4 2023
\(A=\left(\dfrac{x}{x-2}-\dfrac{2}{x+2}\right):\dfrac{x^2+4}{x+2}\)
\(=\left(\dfrac{x.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{x^2-4}{x+2}\)
\(=\left(\dfrac{x^2+2x}{\left(x+2\right)\left(x-2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{x^2-4}{x+2}\)
\(=\left(\dfrac{x^2+2x-2x+4}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{x^2-4}{x+2}\)
\(=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}:\dfrac{x^2-4}{x+2}\)
\(=\dfrac{x^2+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x^2-4}\)
\(=\dfrac{\left(x^2+4\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x^2-4\right)}\)
\(=\dfrac{x^2+4}{\left(x-2\right)\left(x^2-4\right)}\)
TT
3
2
23 tháng 5 2022
Với `x \ne +-2` có:
`M=[x^3]/[x^2-4]-x/[x-2]-2/[x+2]`
`M=[x^3-x(x+2)-2(x-2)]/[(x-2)(x+2)]`
`M=[x^3-x^2-2x-2x+4]/[(x-2)(x+2)]`
`M=[x^3-x^2-4x+4]/[(x-2)(x+2)]`
`M=[x^2(x-1)-4(x-1)]/[x^2-4]`
`M=[(x-1)(x^2-4)]/[x^2-4]`
`M=x-1`
23 tháng 5 2022
\(M=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(=\dfrac{x^3-x\left(x+2\right)-2\left(x-2\right)}{x^2-4}\)
\(=\dfrac{x^3-x^2-2x-2x+4}{x^2+4}=\dfrac{x^3-4x-x^2+4}{x^2-4}=\dfrac{x\left(x^2-4\right)-\left(x^2-4\right)}{x^2-4}\)
\(=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}=x-1\)
23 tháng 12 2020
Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x^2-1\right)-x+2\)
\(=x^3-8-x^3+x-x+2\)
\(=-6\)
Ta có: x + 2 2 x - 4 = x - 2 2 x - 4