ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{-2\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right).2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
A=\(\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
a, Tìm ĐKXĐ
b, Rút gọn
Cho:
\(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)\(-\)\(\dfrac{\sqrt{x}+1}{x-1}\) \(\left(x\ge0,x\ne1\right)\)
a) Rút gọn P
b\()\) CM: P < \(\dfrac{1}{3}\)
Rút gọn D
tính D khi \(\sqrt{3+\sqrt{8}}\)
Tìm x khi D =\(\dfrac{8}{3}\)
D=\(\left(\dfrac{X+1}{X-1}-\dfrac{X-1}{X+1}\right):\left(\dfrac{2}{X^2-1}-\dfrac{X}{X-1}+\dfrac{1}{X+1}\right)\)
Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
P = \(\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{3\left(3\sqrt{x}-5\right)}{x-2\sqrt{x}-3}\)
a) Rút gọn
b) Tìm giá trị của P khi x=\(4+2\sqrt{3}\)
c) Tìm GTNN của P
d) Tìm x khi P>3
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}+3\right)}{x-9}\right)\)\(:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)(với \(x\ge0;x\ne9\))
a) Rút gọn A
b) Tìm x để A<\(-\)1
Cho A=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
Rút gọn A
A=( \(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}-1}\) ) : \(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
a, Tìm ĐKXĐ
b, Rút Gọn A
1`,\(E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)(x>0,x\(\ne\)1)
a,rút gọn E b,Tìm x để E > 0
2,\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right).\left(\sqrt{x}+1\right)\) (x>0,x≠1)
a,rút gọn B b,tìm x để G=2
