Lời giải:
Ta có:
$a^2=3+\sqrt{5+2\sqrt{3}}+3-\sqrt{5+2\sqrt{3}}+2\sqrt{(3+\sqrt{5+2\sqrt{3}})(3-\sqrt{5+2\sqrt{3}})}$
$=6+2\sqrt{3^2-(5+2\sqrt{3})}=6+2\sqrt{4-2\sqrt{3}}=6+2\sqrt{3+1-2\sqrt{3}}$
$=6+2\sqrt{(\sqrt{3}-1)^2}=6+2(\sqrt{3}-1)=4+2\sqrt{3}=(\sqrt{3}+1)^2$
$\Rightarrow a=\sqrt{3}+1$ (do $a\geq 0$)
Do đó:
$a^2-2a-2=4+2\sqrt{3}-2(\sqrt{3}+1)-2=0$ (đpcm)
Rút gọn biểu thức
a) \(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)
b)\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)
c)\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)
d) \(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)
e)\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y},x,y>0\)
f)\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
g)\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}v\text{ới}a>0,a\ne3\)
Mong mng giúp ạ
câu1 rút gọn
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b)\(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(x\ne\sqrt{2},x\ne-\sqrt{2}\right)\)
c)\(\sqrt{9\text{x}^2}-2\text{x}\left(x< 0\right)\)
d)\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
e)\(\dfrac{x^2-5}{x+\sqrt{5}}\left(x\ne-\sqrt{5}\right)\)
\(a,\sqrt{10+\sqrt{9}}\)
b,\(\sqrt{21+6\sqrt{6}}\)-\(\sqrt{21-6\sqrt{6}}\)
c\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
d,\(\sqrt{x+1+2\sqrt{x}}\)
e,\(\sqrt{2\text{x}+3+2\sqrt{x^2+3\text{x}+2}}\)
giúp mik với,cảm ơn trc ạ
Giải giúp em ạ
So sánh A = 2√3 và B= \(\sqrt{4+2\sqrt{3}}\)
A= \(\sqrt{5+2\sqrt{6}}\) và B = \(\dfrac{2\sqrt{2}}{1+2\text{√2}-\sqrt{3}}\)
rút gọn biểu thức
a) A= \(2\sqrt{\frac{1}{2}}+\sqrt{18}\)
b) B= \(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5+3}\right)\)
c) C= \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\left(x>0,x\ne1\right)\)
d) D = \(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x-2}}{x-1}\right)\left(x+\sqrt{x}\right)\left(x>0,x\ne1\right)\)
e) E = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
Bài 1 :Chứng minh các đẳng thức :
a ) \(2\sqrt{2}\left(\sqrt{3}-2\right)\) + \(\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
b ) \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
c ) \(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=6\)
Bài 2 : Rút gọn các biểu thức sau :
a ) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
b ) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
Bài 3 : Rút gọn các biểu thức sau :
a ) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
b ) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
c ) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
d ) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right):\frac{1}{8}\)
1,\(\sqrt{2+\sqrt{\text{x}-5}}=\sqrt{13-x}\)
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
Với a>0, b>0 chứng minh rằng
a)\(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
b)\(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
c)\(\dfrac{\sqrt{a^2+6}}{\sqrt{a^2+5}}>2\)
