\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-2}{2\left(x-1\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)}{2\left(x-1\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\)
√x^2-6x+9=3-x
x^2-1/2x+1/16=x+3/2
√x-2√x-1=√x-1-1
√9-4√5-√5=-2
1: rút gọn \(A=\dfrac{2}{x^2-1}-\dfrac{1}{x^2+x}+\dfrac{x^2-3}{x^3-x}\); \(B=\dfrac{2}{x-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{x^2+6x+2}{x^3-1}\)
2: tìm x: \(\dfrac{4}{3}\left(x-2\right)+\dfrac{\left(x-1\right)\left(x+2\right)}{2}=3-\dfrac{5x\left(1-2x\right)}{4}\)
Gidipt 1) sqrt(x ^ 2 - x) = sqrt(3 - x)
2) sqrt(x ^ 2 - 4x + 3) = x - 2
3) sqrt(4 * (1 - x) ^ 2) - 6 = 0
4) sqrt(x ^ 2 - 4x + 4) = sqrt(4x ^ 2 - 12x + 9)
5) sqrt(x ^ 2 - 4) + sqrt(x ^ 2 + 4x + 4) = 0
6) 1sqrt(x + 2sqrt(x - 1)) + sqrt(x - 2sqrt(x - 1)) = 2
cho A =((√x-2)/(x-1)-(√x+2)/(x+2*√x+1))*((x^2-2*x+1)/2) chứng minh rằng 0<x<1 thì A>0
Tìm x, biết:
a, \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)
b, \(\sqrt{1-12x+36x^2}=5\)
c, \(\sqrt{x+2\sqrt{x-1}}=2\)
Cho biểu thức \(A=\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^2}-1+x}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
Tính A khi \(x=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
Tìm điều kiện xác định và rút gọn biểu thức
a) Q= (\(\frac{1}{x-1}\)-\(\frac{1}{x}\)) :(\(\frac{x+1}{x-2}\)-\(\frac{x+2}{x-1}\))
b) C= (\(\frac{x+2}{x^2-x}\)+\(\frac{x-2}{x^2+x}\)) . \(\frac{x^2-1}{x^2+2}\)
1. A= \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Chứng minh: A<1
cho biểu thưc P=\(\left(\frac{2x^2+1}{x^2-1}-\frac{x}{x^2+x+1}\right).\left(\frac{x^3+1}{1+x}-x\right)\)
A = \(\frac{x^2+2x}{x^2-2x+1}\)\(:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
a, Rút gọn A
b, Tìm GTNN của A khi x >1
