\(x-2\sqrt{x}=0\left(x\ge0\right)\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
,giải pt
a,\(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
b,\(\sqrt{x^2-1}-x^2+1=0\)
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:
a, A = \(\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\) (đk: x ≥ 0 và x ≠ 36)
b, B = \(\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\) (đk: x ≥ 0 và x ≠ 9)
c, C = \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2\) (đk: a > 0, b > 0 và a ≠ b)
d, D = \(\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\) (đk: a ≥ 0, a ≠ 2, a ≠ 4)
A= x-2 2+ sqrt x (x>=0); ==( 8x sqrt x -1 2x- sqrt x - 8x sqrt x +1 2x+ sqrt x )= 2x+1 2x-1 vdi x>0,x ne 1 2 ;x ne- 1 2 MS05. Cho A =- a. Rút gọn B. b. Tim x d hat e A B =1
bài 1: rút gọn
A= \(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) X>=0; x#0
B= \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\left(\dfrac{2}{x^2-2x+1}\right)\) x>=0; x\(\ne\)1
bài 2:
\(\dfrac{1}{2x-3\sqrt{x}+2}\) x>=0
Tìm GTNN của A
b)\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)+\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)+\(\dfrac{5\sqrt{x}+2}{4-x}\)( điêu kiện:x≥0;x≠4)
Rút gọn
\(\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}\left(a>0,x>0\right)\)
1/Rút gọn
A=\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3+x}\right)}\)(x>0; y>0; x#y)
B= \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)( x>0)
C=\(\left(\frac{x+1}{\sqrt{x}}+2\right).\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x\sqrt{x}+1\right)}\)(x>0)
D=\(\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right):\left(x-1\right)-\frac{2}{\sqrt{x}-1}\)(x>=0; x#1)
giúp em với ạ em đang cần gấp ạ
\(6x^2+2x+\sqrt[3]{3x^2+x+4}-10=0\)
\(x+1+\sqrt{x^24x+1}=3\sqrt{x}\)
\(x^2+2x\sqrt{x^2+4x+1}=3\sqrt{x}\)
\(\sqrt{x+8}+\dfrac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
Giải phương trình :
a, \(\sqrt{x+1}=x-1\)
b, \(x-\sqrt{2x+3}=0\)
c, \(\sqrt{x-2}-3\sqrt{\left(x-2\right)\left(x+2\right)}=0\)
d, \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)
e, \(2\sqrt{x+3}=9x^2-x-4\)
f, \(\sqrt{x+1}-\sqrt{x-7}=\sqrt{12-x}\)
g, \(\sqrt{2x+5}-\sqrt{3x-5}=2\)
h, \(\sqrt{x}-\sqrt{x-1}-\sqrt{x-4}+\sqrt{x+9}=0\)
i, \(x^2+2x-\sqrt{x^2+2x+1}-5=0\)
k, \(\sqrt{x+8-6\sqrt{x+1}}=4\)
l, \(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)
Làm được phần nào thì giúp mình nha đang cần gấp !!!
