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a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
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
bài 1:
a)\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)\(do2>\sqrt{3}\)
\(=6+3\sqrt{3}-2\sqrt{2}-\sqrt{6}\)
b) \(\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)do\sqrt{5}>\sqrt{2}\)
\(=\sqrt{15}-\sqrt{6}+5-\sqrt{10}\)
c)\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\)
\(=\left(2+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)do\sqrt{5}>2\)
\(=5-4\)
\(=1\left(hđt.3\right)\)
d)\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)do\sqrt{5}>\sqrt{3}\)
\(=5-3\)
\(=2\)
e)\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\)
\(=2\left(2-4+9\right)\)
\(=2.7=14\)
f)\(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)
\(=2-\sqrt{6-2\sqrt{5}}\)
\(=2-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2-\left(\sqrt{5}-1\right)\)
\(=2-\sqrt{5}+1\)
\(=3-\sqrt{5}\)
g)\(\sqrt{3}-\sqrt{2}\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\sqrt{3}-\sqrt{6}-2\)
h) \(\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right)\sqrt{2}+2\sqrt{5}\)
\(=\left(2-\sqrt{6+2\sqrt{5}}\right)+2\sqrt{5}\)
\(=\left(2-\sqrt{\left(\sqrt{5}+1\right)^2}\right)+2\sqrt{5}\)
\(=2-\left(\sqrt{5}+1\right)+2\sqrt{5}\left(do\sqrt{5}>1\right)\)
\(=2-\sqrt{5}-1+2\sqrt{5}\)
\(=1-\sqrt{5}\)
bài 2)
a) \(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)
\(\Leftrightarrow2x-1=5\)hoặc \(\Leftrightarrow2x-1=-5\)
\(\Leftrightarrow x=3\)hoặc \(\Leftrightarrow x=-2\)
Vậy x = 3 hoặc x = -2
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1)
a)\(\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2=x-y\)
b)\(\left(\sqrt{x}-3\right).\left(\sqrt{x}+2\right)=\left(\sqrt{x}\right)^2+2\sqrt{x}-3\sqrt{x}-6=x-\sqrt{x}-6\)
c)\(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-\left(\sqrt{5}\right)^2\)
=\(4-5=1\)
d)\(\sqrt{\left(-5\right)^2.3^2}=\left(5\right).3=15\)
e)\(\sqrt{\frac{5}{27}.\frac{8}{20}}=\sqrt{\frac{2}{27}}\)
ĐÂy này nhớ **** vài câu nha
a, \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{2}\sqrt{5-\sqrt{21}}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{7-2.\sqrt{3}.\sqrt{7}+3}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\)