Lời giải:
$\sqrt{27}+\sqrt{96}-\sqrt{150}-\sqrt{12}$
$=3\sqrt{3}+4\sqrt{6}-5\sqrt{6}-2\sqrt{3}=\sqrt{3}-\sqrt{6}$
$=\sqrt{3}(1-\sqrt{2})$
Do đó: $(\sqrt{27}+\sqrt{96}-\sqrt{150}-\sqrt{12}):(1-\sqrt{2})=\sqrt{3}$
Rút gọn:
\(2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
rút gọn
a) \(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)
b) \(\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
c) \(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)\)
Rút gọn:
\(A=\left(\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x+\sqrt{x}}\right).\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}-1\)
Rút gọn:
\(A=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right).\dfrac{\sqrt{a}+1}{\sqrt{a}}\left(a>0,a\ne1\right)\)
1,Rút gọn
a, \(\left(2\sqrt{2}-1\right)\left(\sqrt{8}+1\right)\)
b, \(\left(\sqrt{12}+\sqrt{75}-2\sqrt{27}\right):\sqrt{3}\)
c, \(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
d, \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
e, \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
f, \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}+3\right)}{x-9}\right)\)\(:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)(với \(x\ge0;x\ne9\))
a) Rút gọn A
b) Tìm x để A<\(-\)1
1`,\(E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)(x>0,x\(\ne\)1)
a,rút gọn E b,Tìm x để E > 0
2,\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right).\left(\sqrt{x}+1\right)\) (x>0,x≠1)
a,rút gọn B b,tìm x để G=2
Rút gọn các biểu thức sau :
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
b) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\)
Rút gọn biểu thức với a>0: A=\(\dfrac{a^2-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}\)
