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câu g
(câu cuối) đề nhiều trôi hết nhìn thấy mỗi câu (g)
\(G=0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(G=0,1.10\sqrt{2}+\dfrac{2.2}{10}\sqrt{2}+0,4.5\sqrt{2}\)
\(G=\sqrt{2}\left(1+\dfrac{2}{5}+2\right)=\dfrac{\sqrt{2}\left(5+2+10\right)}{5}=\dfrac{17\sqrt{2}}{5}\)
a) \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
= \(\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
= \(-16\sqrt{3}\)
b) \(\left(a.\sqrt{\dfrac{a}{b}}+2\sqrt{ab}+b.\sqrt{\dfrac{b}{a}}\right)\sqrt{\dfrac{a}{b}}\)
= \(\dfrac{a^2}{b}+2a+b\) = \(\dfrac{a^2+\left(2a+b\right)b}{b}\) = \(\dfrac{a^2+2ab+b^2}{b}\) = \(\dfrac{\left(a+b\right)^2}{b}\)
c) \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\) = \(3+2-5=0\)
d) \(3+\sqrt{18}+\sqrt{3}+\sqrt{8}\) = \(3+3\sqrt{2}+\sqrt{3}+2\sqrt{2}\)
= \(3+\sqrt{3}+5\sqrt{2}\)
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)
b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)
c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)
d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)
f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}\)
\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)
= - 1
\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}+1}{2}\)
\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)
\(=\dfrac{2\sqrt{2}+2}{\sqrt{3+2\sqrt{2}}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
= 2
\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)
\(=4+9+16+49\)
= 78
\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)
\(=\sqrt{x}-\sqrt{y}\)
\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)
\(=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}\)
\(=2-\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)-1\)
= 3
Ta có:
\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)
\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)
\(=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}\)
\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)
\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{2}\right)^2}{\left(\sqrt{a}-\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)
Bài 50:
\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)
\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
a: \(=2\sqrt{2}+30\sqrt{2}-3\sqrt{2}+6\sqrt{2}=26\sqrt{2}\)
b: \(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}+\sqrt{3}+\dfrac{5}{2}\sqrt{3}=-\dfrac{9}{2}\sqrt{3}\)
\(a.\sqrt{54}:\sqrt{2}:\sqrt{3}\\ =\sqrt{54:2:3}=\sqrt{9}=3\\ b.\dfrac{\sqrt{3}}{\sqrt{75}}:\dfrac{\sqrt{52}}{\sqrt{117}}=\sqrt{\dfrac{1}{25}}:\sqrt{\dfrac{4}{9}}\\ =\dfrac{1}{5}:\dfrac{2}{3}=\dfrac{1}{5}\cdot\dfrac{3}{2}=\dfrac{3}{10}\)