a: =>2x^2-3x-2=6x+2
=>2x^2-9x-4=0
=>\(x=\dfrac{9\pm\sqrt{113}}{4}\)
b: \(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a) \(\dfrac{2x^2-3x-2}{2x+1}\)=3
b)B=\(\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}\)=\(\dfrac{1}{\sqrt{x}-2}\)
Tìm điều kiện để các biểu thức sau xác định
a)\(\sqrt{x+1}-\dfrac{1}{2}\)
b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)
c)\(\sqrt{x+1}-\sqrt{x-2}\)
d)\(\sqrt{2-3x}-\sqrt{1-2x}\)
e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)
f)\(\dfrac{1}{2}.\sqrt{x-\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\sqrt{x}-1}\)
g)\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\)
h)\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x^2+2}}\)
Tìm x
1) \(\sqrt{\dfrac{3x-1}{x+2}}=2\)
2)\(\sqrt{\dfrac{5x-7}{2x- 1}}=2\)
3)\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
4) \(\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Tìm x
a)\(\sqrt{2x-1}=3\)
b)\(\sqrt{1-3x}=\dfrac{1}{2}\)
c)\(\sqrt{\left(x-1\right)^2}=\dfrac{1}{2}\)
d)\(\sqrt{\left(1+2x\right)^2}=\dfrac{\sqrt{3}}{2}\)
e)\(\sqrt{\left(1-2x\right)^2=|x-1|}\)
Tìm `ĐKXĐ`:
\(\sqrt{\dfrac{-5}{6+x}}\)
\(\sqrt{\dfrac{-2}{6-x}}\)
\(\sqrt{\dfrac{-x+3}{-6}}\)
\(\sqrt{\dfrac{7x-1}{-9}}\)
\(\sqrt{\dfrac{x+2}{x^2+2x+1}}\)
\(\sqrt{\dfrac{x-2}{x^2-2x+4}}\)
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
Tìm điều kiện có nghĩa:
1) \(-\dfrac{1}{\sqrt{a+2}}\)
2) \(\sqrt{\dfrac{3}{\left(x-2\right)^2}}\)
3) \(\sqrt{\dfrac{-3}{a^2-4a+4}}\)
4) \(\sqrt{\dfrac{2}{x^2+2x+2}}\)
5) \(\sqrt{\dfrac{-3}{x^2-4x+5}}\)
6) \(\sqrt{\dfrac{-4}{x^2-1}}\)
7) \(\sqrt{\dfrac{x+1}{x-2}}\)
8) \(\sqrt{\dfrac{x-2}{x+3}}\)
Chứng minh đẳng thức
a. \(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}1.\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x-1}{x}=\dfrac{2x}{x-1}\)
b. \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
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
