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A= \(\dfrac{x}{x+2}\)+\(\dfrac{2x}{x-2}\)+\(\dfrac{2\left(x^2-x-2\right)}{4-x^2}\)
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AH
Akai Haruma
Giáo viên
2 tháng 3 2021
Lời giải:
ĐK: $x\geq 0; x\neq 1$
\(A=\left[\frac{(\sqrt{x}-1)(x+2\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}+2)}\right].\frac{\sqrt{x}-1}{(\sqrt{x}-1)(2\sqrt{x}+3)}\)
\(=\left(\frac{x+2\sqrt{x}+2}{\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}+3}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}.\frac{1}{2\sqrt{x}+3}=\frac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(2\sqrt{x}+3)}=\frac{\sqrt{x}+1}{2\sqrt{x}+3}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2022
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
A
1
NH
5 tháng 2 2023
\(\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}-\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9}{x\left(x-3\right)}\right)\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ =\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6}{x}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-3}{x-1}\)
4 tháng 4 2021
Ta có: \(A=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}+\dfrac{2x+4}{4-x^2}\right)\cdot\left(x+\dfrac{5}{x-3}\right)\)
\(=\left(\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\left(\dfrac{x\left(x-3\right)+5}{\left(x-3\right)}\right)\)
\(=\dfrac{x^2-4x+4+x^2+2x-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x^2-4x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x\left(x^2-3x+5\right)}{\left(x+2\right)\left(x-3\right)}\)
7 tháng 7 2021
\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)
\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)
b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)
6 tháng 7 2021
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
\(A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2-2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+2x^2+4x-2x^2+2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{x-2}\)