Lời giải:
ĐKXĐ: $x^2\geq 2$
PT $\Leftrightarrow (x^2-2)-\sqrt{x^2-2}-2=0$
$\Leftrightarrow a^2-a-2=0$ (đặt $\sqrt{x^2-2}=a, a\geq 0$)
$\Leftrightarrow (a+1)(a-2)=0$
$\Leftrightarrow a=-1$ hoặc $a=2$.
Do $a\geq 0$ nên $a=2$
$\Leftrightarrow \sqrt{x^2-2}=2$
$\Leftrightarrow x^2-2=4$
$\Leftrightarrow x^2=6\Leftrightarrow x=\pm \sqrt{6}$ (thỏa mãn)
1) x-\(7\sqrt{x-3}\) -9=0 2) \(\sqrt{x+3}\) =5-\(\sqrt{x-2}\) 3) \(\sqrt{x-4\sqrt{x+4}}\) =3 4) \(\sqrt{8-\dfrac{2}{3}x}-5\sqrt{2}\) =0 5) \(\sqrt{x^2-4x+4}\) =2-x
bài 1 : giải phương trình:
a. \(\sqrt{x+2\sqrt{ }x-1}=2\)
b. \(\sqrt{x^2-4x+4}=\sqrt{4x^212x+9}\)
c.\(\sqrt{x+4\sqrt{ }x-4}=2\)
d. \(\sqrt{x^2-6x+9}=2\)
e. \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
f. \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
Cho 0<x<2. Chứng minh rằng:
\(\dfrac{4-\sqrt{4-x^2}}{\sqrt{\left(2+x\right)^3}+\sqrt{\left(2-x\right)^3}}\) + \(\dfrac{4+\sqrt{4-x^2}}{\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}}\) = \(\dfrac{\sqrt{2+x}}{x}\)
Giải các phương trình sau :
1/\(\sqrt{x+2+4\sqrt{x-2}}=5\)
2/\(\sqrt{x+3+4\sqrt{x-1}}=2\)
3/\(\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\)
4/\(\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\)
i) \(2\sqrt{x+2+2\sqrt{x+1}}-\sqrt{x+1}=4\)
ii)\(\sqrt{x+4}+\sqrt{x-4}=2x-12+2\sqrt{x^2-16}\)
Rút gọn biểu thức :
\(D=\frac{x+2+\sqrt{x^2+4}}{x+2-\sqrt{x^2-4}}+\frac{x+2-\sqrt{x^2}-4}{x+2+\sqrt{x^2}-4}\)
\(Q=\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)
\(\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2}-4}+\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2}-4}\)
giải phương trình sau?
1)\(\sqrt{x+1}+\sqrt{x+10}=\sqrt{x+5}+\sqrt{x+2}\)
2) \(8\sqrt{x^3+1}=3\left(x^2-2x\right)\)
3) \(20\sqrt{\frac{x-2}{x+1}}-5\sqrt{\frac{x+2}{x-1}}=-4\sqrt[4]{\frac{x^2-4}{x^2-1}}\)
4)\(\sqrt{x^2+x-1}+\sqrt{-x^2+x+1}=x^2-x-2\)
5) \(\frac{4x^2}{\sqrt{x^4+x}}=-x^2+4x-3\)
6)\(\sqrt[4]{x}+\sqrt[4]{2-x}=2\)
Rút gọn:
\(A=\sqrt{x+\sqrt{x^2-4}}+\sqrt{x-\sqrt{x^2-4}}\)
\(B=\sqrt{10x-6\sqrt{x^2-2x}-2}+\sqrt{5x+4\sqrt{x^2-2x}-2}\)
\(C=\frac{\sqrt{2+\sqrt{-x^2+6x-8}}}{x-3}\)
\(D=\sqrt{\frac{17}{4}+2\sqrt{4-x^2}+\sqrt{4+2\sqrt{4-x^2}}}\)
Giúp mình với các bạn
a) \(8\sqrt{x+2}\) + \(\sqrt{11-x}\) - \(2\sqrt{22+9x-x^2}\)+ 4 =0
b) \(\sqrt{1+4x}\)+ \(2\sqrt{2-x}\)+\(2\sqrt{\left(1+4x\right)\left(2-x\right)}\)=3
c) \(\sqrt{8+\sqrt{x}}\)+\(\sqrt{5-\sqrt{x}}\)=5
d) \(\sqrt{x^4-1}\)-2 =\(\sqrt{x-1}\)- \(2\sqrt{x^3+x^2+x+1}\)
