Với `x > 0,x \ne 1` có:
`(1/[x-\qrt{x}]+1/[\sqrt{x}-1]):[\sqrt{x}+1]/[x\sqrt{x}-2x+\sqrt{x}]`
`=[1+\sqrt{x}]/[\sqrt{x}(\sqrt{x}-1)].[\sqrt{x}(\sqrt{x}-1)^2]/[\sqrt{x}+1]`
`=\sqrt{x}-1`
Cho P= \((\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{\sqrt{x}}):(\dfrac{2x+\sqrt{x}-1}{\sqrt{x}-x\sqrt{x}}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\sqrt{x}+x^{2}})\)
a) Rút gọn P
b) so sánh P với \(\dfrac{3}{4}\).
c) tìm x để P=1
Rút gọn:
1) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\) với x >1
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\) với x ≠ 4, x ≠16, x >0
Rút gọn:
\(\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
Rút gọn
(\(\dfrac{\sqrt{x}}{3+\sqrt{x}}\)+\(\dfrac{2x}{9-x}\)):(\(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\))
(\(\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}+\dfrac{x+9}{25-x}\)):\(\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)
(\(\dfrac{1}{x-4}-\dfrac{1}{x-4\sqrt{x}+4}\)):\(\dfrac{\sqrt{x}}{2\sqrt{x}-x}\)
Rút gọn biểu thức:
1, \(B=\left(\dfrac{x.\sqrt{x}+x+\sqrt{x}}{x.\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right).\dfrac{x-1}{2x+\sqrt{x}-1}\)với x>-0, x khác 1, x khác \(\dfrac{1}{4}\)
2, \(A=\dfrac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\) với x\(\ge\)0:x\(\ne\)0
Rút gọn biểu thức
\(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right).\dfrac{x-6\sqrt{x}+9}{\sqrt{x}}\)
\(\dfrac{x}{\sqrt{x}+1}+\dfrac{2x+\sqrt{x}}{x+\sqrt{x}}\)
Rút gọn: \(1-\left(\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{ }}x\right)\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\)
cho A=\(\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x\sqrt{x-x}}{1-\sqrt{x}}\)
rút gọn biểu thức A
giúp mình với mình cảm ơn
Rút gọn:
1) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
2)\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
3) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
4) \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
Mng giúp e vs ạ, cần gấp :<
\(A=\dfrac{2x+2\sqrt{x}+2}{x-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Rút gọn \(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\)
