ĐKXĐ: \(x\ge3\)
Bình phương 2 vế:
\(x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=3\end{matrix}\right.\)
\(\sqrt{28-6\sqrt{3}}\)
\(\sqrt{6-\sqrt{20}}\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\cdot\left(x+2\right)}}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)
A=(\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\))/(\(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\))
Tìm x
\(a.\sqrt{2+\sqrt{3+\sqrt{x}}=3}\)
\(b.\sqrt{x^2-4}+\sqrt{x+2}=0\)
\(c.\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
rút gọn
\(\dfrac{9-x}{\sqrt{x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\) (với x>_9)
\(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)/\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) (với x>=0, x#1)
\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) ( với x>_6)
\(\sqrt{m^2+6m+9}+\sqrt{m^2-6m+9}\) (m bát kì)
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\dfrac{x+1}{\sqrt{x}}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}/\dfrac{\sqrt{x}-\sqrt{y}}{x-y}\)
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right)/\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}\)
Giải các phương trình sau:
a)\(\sqrt[3]{9-x}+\sqrt[3]{7+x}=4\)
b)\(\sqrt{x-1}\cdot\sqrt[4]{x^2-4}=\sqrt{x-2}\cdot\sqrt[4]{x^2-1}\)
c)\(\sqrt[4]{9-x^2}+\sqrt{x^2-1}-2\sqrt{2}=\sqrt[6]{x-3}\)
\(a,x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(b,\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
c, \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{8-6\sqrt{x-1}+x}=1\)
d, \(\sqrt{3-x+x^2}-\sqrt{2+x-x^2=1}\)
e,\(\sqrt{4x-9}+2\sqrt{3x^2-5x+2}=\sqrt{3x-2}+\sqrt{x-1}\)
1) Tính
a) \(2\sqrt{24}-9\sqrt{\dfrac{2}{3}}+\dfrac{\sqrt{6}-6}{\sqrt{6}}\)
b) \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right)\): \(\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
Bài 1: Cho biểu thức A = 1 - \(\dfrac{\sqrt{x}}{1+\sqrt{x}}\), B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)+ \(\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\)- \(\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)
(với x ≥ 0, x ≠ 4, x ≠ 9)
a, Tính giá trị của A biết x = 6-2\(\sqrt{5}\)
b, Rút gọn P = A : B
c, Tìm giá trị nhỏ nhất của P
Giải pt:
a) \(\sqrt{2x^2-3}\)=\(\sqrt{4x-3}\)
b) \(\sqrt{2x-1}\)=\(\sqrt{x-1}\)
c) \(\sqrt{x^2-x-6}\)=\(\sqrt{x-3}\)
d) \(\sqrt{x^2-x}\)=\(\sqrt{3x-5}\)
Giúp em với, anh thịnh giúp em xíu á
giải pt sau
1, \(\sqrt{5-2x}=6\)
2,\(\sqrt{2-x}-\sqrt{x+1}=0\)
3, \(\sqrt{4x^2+4x+1}=6\)
4,\(\sqrt{x^2-10x+25}=x-2\)
Giair phương trình
\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-1}+\sqrt{x^2+x-6}\)
