rút gon B=2/√x+1+√x-15/25-x
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15 tháng 6 2023
a: \(A=\sqrt{x-2\sqrt{x}+1}=\left|\sqrt{x}-1\right|\)
Khi x=25 thì A=|5-1|=4
b: \(B=\dfrac{-x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{-x+2\sqrt{x}}{x-4}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}\)
VD
0
PB
1
7 tháng 12 2017
= 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 + .... + 1/x+2016 - 1/x+2017
= 2/x+1 - 1/x+2017
k mk nha
16 tháng 10 2021
\(=\dfrac{x+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
19 tháng 12 2020
a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)
\(=\left(\dfrac{x}{3\left(x-3\right)}-\dfrac{2x-3}{x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\left(\dfrac{x^2}{3x\left(x-3\right)}-\dfrac{3\left(2x-3\right)}{3x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{x^2+6x+9}\)
b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)
\(=\dfrac{-6}{x-2}\)
19 tháng 7 2021
Ta có: \(G=\left(\dfrac{x-\sqrt{x}+2}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x+2\sqrt{x}+1}{2x-2\sqrt{x}}\)
\(=\dfrac{x-\sqrt{x}+2-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
FT
1
6 tháng 12 2017
ta có B=2+\(\frac{x}{x-2}\)- \(\frac{4x^2}{x^2-4}\)- \(\frac{2-x}{x+2}\)
=\(\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{2\left(x-2\right)\left(x+2\right)+x\left(x+2\right)-4x^2-\left(x-2\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)\left\{2\left(x-2\right)+x\right\}-\left\{4x^2-\left(x-2\right)^2\right\}}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(3x-4\right)-\left(2x-x+2\right)\left(2x+x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)\left(3x-4\right)-\left(x+2\right)\left(3x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2\right)\left(3x-4-3x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-2}{x-2}\)
Sửa đề: \(B=\dfrac{2}{\sqrt{x}+5}-\dfrac{\sqrt{x}-15}{x-25}\)
\(=\dfrac{2\sqrt{x}-10-\sqrt{x}+15}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{1}{\sqrt{x}-5}\)
\(B=\dfrac{2}{\sqrt{x}+1}+\dfrac{\sqrt{x}-15}{25-x}\left(x\ge0;x\ne25\right)\\ B=\dfrac{2\left(25-x\right)+\left(\sqrt{x}-15\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(25-x\right)}\\ B=\dfrac{50-2x+x-14\sqrt{x}-15}{\left(\sqrt{x}+1\right)\left(25-x\right)}\\ B=\dfrac{-x-14\sqrt{x}+35}{\left(\sqrt{x}+1\right)\left(25-x\right)}\)