\(\sqrt{\dfrac{1}{x^2-2x+1}}\)
= \(\dfrac{\sqrt{1}}{\sqrt{\left(x-1\right)^2}}\)
= \(\dfrac{1}{|x-1|}\)
= \(\left[{}\begin{matrix}\dfrac{1}{x-1}\\\dfrac{1}{-\left(x-1\right)}\end{matrix}\right.\)
= \(\left[{}\begin{matrix}\dfrac{1}{x-1}\\\dfrac{-1}{x-1}\end{matrix}\right.\)
Tìm Min và Max(nếu có)
A=2x-\(\sqrt{x}\)
B=x+\(\sqrt{x}\)
C=1+\(\sqrt{2-x}\)
D=\(\sqrt{-x^2+2x+5}\)
E=\(\dfrac{1}{2x-\sqrt{x}+3}\)
F=\(\dfrac{1}{3-\sqrt{1-x^2}}\)
Rút gọn biểu thức
\(a.\dfrac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
\(b.x\sqrt{2x+2}+\left(x+1\right)\sqrt{\dfrac{2}{x+1}}-4\sqrt{\dfrac{x+1}{2}}\)
- Rút gọn :
a) \(A=\dfrac{\sqrt{x+\sqrt{2x-1}}}{\sqrt{2}}\)
b) \(B=\dfrac{1}{x+\sqrt{1+x^2}}+\dfrac{1}{x-\sqrt{1+x^2}}+2\)
\(\dfrac{1+2x}{1+\sqrt{1+2x}}\)+\(\dfrac{1-2x}{1-\sqrt{1-2x}}\) với x=\(\dfrac{\sqrt{3}}{4}\)
Tính giá trị biểu thức trên..
Giải phương trình:
a) \(\dfrac{1}{x+\sqrt{1+x^2}}+\dfrac{1}{x-\sqrt{1+x^2}}+2=0\)
b) \(2x-5a\sqrt{x-a}+2a\left(a-1\right)=0\) với a>0
Rút gọn biểu thức : \(x\sqrt{2x+2}+\left(x+1\right)\sqrt{\dfrac{2}{x+1}}-4\sqrt{\dfrac{x+1}{2}}\)
Bài 1: Tính:
\(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
Bài 2: Rút gọn rồi tính:
a) A=\(\dfrac{a^4-4a^2+3}{a^4-12a^2+27},a=\sqrt{3}-\sqrt{2}\)
b) \(B=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}},h=3\)
c) \(C=\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}x+2},x=2\left(\sqrt{3}+1\right)\)
d) \(D=\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{3}{\sqrt{1-a^2}}+1\right),a=\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
Mọi người giúp em với!!!!!!!!!!!!!!
khử mẫu bt lấy căn :
a) \(3xy\cdot\sqrt{\dfrac{2}{xy}}\)
b)\(x\cdot\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}\)
c) \(xy\cdot\sqrt{\dfrac{1}{xy}}+x\cdot\sqrt{\dfrac{y}{x}}-y\cdot\sqrt{\dfrac{x}{y}}\)
trục căn thức
a) \(\dfrac{1}{\sqrt{x-1}};\dfrac{a+2}{\sqrt{a^2-4}};\dfrac{x-y}{\sqrt{x^2-y^2}};\dfrac{a}{\sqrt{x^2}}\) (n lẻ)
b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}\)
c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}\)
