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Câu b nhé:
Ta có:
\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)
So Sánh
a.\(\dfrac{1}{4}\sqrt{8}\) và \(\dfrac{2}{3}\sqrt{12}\)
Có:\(\dfrac{1}{4}\sqrt{8}\) và \(\dfrac{2}{3}\sqrt{12}\)
= \(\dfrac{1}{4}.2\sqrt{2}\) và \(\dfrac{2}{3}.2\sqrt{3}\)
=\(\dfrac{\sqrt{2}}{2}\)và \(\dfrac{4\sqrt{3}}{3}\)
=> \(\dfrac{1}{4}\sqrt{8}< \dfrac{2}{3}\sqrt{12}\)
b. \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}\)và \(6\sqrt{\dfrac{1}{35}}\)
Có \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}\) và \(6\sqrt{\dfrac{1}{35}}\)
=\(\dfrac{5}{2}.\dfrac{\sqrt{6}}{6}\) và \(6.\dfrac{\sqrt{35}}{35}\)
=\(\dfrac{5\sqrt{6}}{12}\) và \(\dfrac{6\sqrt{35}}{35}\)
=> \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{35}}\)
c. \(\dfrac{1}{6}\sqrt{18}\) và \(\dfrac{1}{2}\sqrt{2}\)
=\(\dfrac{1}{6}.3\sqrt{2}\) và \(\dfrac{1}{2}\sqrt{2}\)
=\(\dfrac{\sqrt{2}}{2}\) và \(\dfrac{\sqrt{2}}{2}\)
=> \(\dfrac{1}{6}\sqrt{18}=\dfrac{1}{2}\sqrt{2}\)
a,\(\dfrac{1}{4}\sqrt{8}=\dfrac{1}{\sqrt{2}}\)
\(\dfrac{2}{3}\sqrt{12}=\dfrac{4}{\sqrt{3}}\)
=> \(\dfrac{1}{4}\sqrt{8}< \dfrac{2}{3}\sqrt{12}\)
a/ \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=7-4\sqrt{3}+7+4\sqrt{3}=14\)
a) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)
b) \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}+2}-\dfrac{12}{3-\sqrt{6}}=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}+2}\right)-\dfrac{12}{3-\sqrt{6}}\)
\(=\left(\dfrac{15\left(\sqrt{6}+2\right)+4\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}+2\right)}\right)-\dfrac{12}{3-\sqrt{6}}=\dfrac{15\sqrt{6}+30+4\sqrt{6}+4}{6+2\sqrt{6}+\sqrt{6}+2}-\dfrac{12}{3-\sqrt{6}}\) \(=\dfrac{34+19\sqrt{6}}{8+3\sqrt{6}}-\dfrac{12}{3-\sqrt{6}}=\dfrac{\left(34+19\sqrt{6}\right)\left(3-\sqrt{6}\right)-12\left(8+3\sqrt{6}\right)}{\left(8+3\sqrt{6}\right)\left(3-\sqrt{6}\right)}\)
\(=\dfrac{102-34\sqrt{6}+57\sqrt{6}-114-96-36\sqrt{6}}{24-8\sqrt{6}+9\sqrt{6}-18}=\dfrac{-108-13\sqrt{6}}{6+\sqrt{6}}\)
c) \(\sqrt{2+\sqrt{3}}+\sqrt{2+\sqrt{3}}=2\sqrt{2+\sqrt{3}}=\sqrt{2}.\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
câu này mk cảm thấy đề sai thì phải ; mà nếu o phải đề sai thì lời giải đó nha
a. Ta có \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
Vậy \(3\sqrt{3}>\sqrt{12}\)
b. Ta có \(7=\sqrt{49}\), \(3\sqrt{5}=\sqrt{45}\)
Vì \(\sqrt{49}>\sqrt{45}\)nên \(7>3\sqrt{5}\)
c. Ta có \(\dfrac{1}{3}\sqrt{51}=\dfrac{\sqrt{51}}{3}\), \(\dfrac{1}{5}\sqrt{150}=\sqrt{6}=\dfrac{3\sqrt{6}}{3}=\dfrac{\sqrt{54}}{3}\)
Vì \(\dfrac{\sqrt{51}}{3}< \dfrac{\sqrt{54}}{3}\) nên \(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
d. Ta có \(\dfrac{1}{2}\sqrt{6}=\dfrac{\sqrt{6}}{2}\), \(6\sqrt{\dfrac{1}{2}}=3\sqrt{2}=\dfrac{6\sqrt{2}}{2}\)
Vì \(\dfrac{\sqrt{6}}{2}< \dfrac{6\sqrt{2}}{2}\Rightarrow\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a)\(=\left(\sqrt{2}\cdot\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{\sqrt{3}+2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{2}\cdot\sqrt{\sqrt{3}+2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{2\cdot\left(\sqrt{3}+2\right)}\)
=\(\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{2\sqrt{3}+4}\)
=\(\left(3-2\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)
=\(\left|\sqrt{3}+1\right|\)=\(\sqrt{3}+1\)
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
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
Ta có: \(\sqrt{150}-6\sqrt{\dfrac{1}{6}}-12\cdot\sqrt{\dfrac{1}{24}}-\dfrac{18}{\sqrt{6}}\)
\(=5\sqrt{6}-\sqrt{6}-\sqrt{6}-3\sqrt{6}\)
=0