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\(a.\dfrac{\sqrt{8}+2}{\sqrt{2}+1}.\sqrt{2}=\dfrac{\sqrt{2}.2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\sqrt{2}\)
\(b.\dfrac{5+\sqrt{5}}{\sqrt{5}+1}.\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}.\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\sqrt{5}.\sqrt{5}=5\)
\(c.\dfrac{2}{\sqrt{2}}+\sqrt{18}+\sqrt{32}=\sqrt{2}+\sqrt{9.2}+\sqrt{16.2}=\sqrt{2}+3\sqrt{2}+4\sqrt{2}=8\sqrt{2}\)
\(d.\left(1-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(1+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\right)=\left(1-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(1+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\right)=\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)=1-2=-1\)
a) \(2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}=2\sqrt{2^2.5}-\sqrt{5^2.2}+3\sqrt{4^2.5}-\sqrt{8^2.5}\\ =4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)
b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=\sqrt{4^2.2}-\sqrt{5^2.2}+\sqrt{3^2.2}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
c) \(3\sqrt{3}+4\sqrt{2}-5\sqrt{27}=3\sqrt{3}+4\sqrt{2}-5\sqrt{3^2.3}=3\sqrt{3}+4\sqrt{2}-15\sqrt{3}=4\sqrt{2}-12\sqrt{3}\)
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}\\ =\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\left(\sqrt{3+1}\right)^2-1^2}\\ =\dfrac{2\sqrt{3}}{\sqrt{3}}=2\)
e)\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=2^2-\left(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)^2=4-\left(\dfrac{9+6\sqrt{3}+3}{3+2\sqrt{3}+1}\right)\\ =4-\left(\dfrac{6\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}\right)=4-3=1\)
b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=\left(4-5+3\right)\sqrt{2}=2\sqrt{2}\)
a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)
b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)
c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)
d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)
f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a: \(B=\dfrac{12+2\sqrt{35}+12-2\sqrt{35}}{2}-\dfrac{5}{2}\)
=12-5/2
=9,5
b: \(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+4-2\sqrt{5}+4}{1}=8\)
a: \(\dfrac{\sqrt{50}-\sqrt{32}+\sqrt{8}}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}-4\sqrt{2}+2\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{\sqrt{2}}=3\)
b: \(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
\(=\dfrac{4\left(\sqrt{5}+1\right)}{5-1}-\sqrt{5}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1