\(\dfrac{1}{2}\times\sqrt{100}-\sqrt{\dfrac{1}{16}}+\left(\dfrac{1}{3}\right)^0\)
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23 tháng 8 2021
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
5 tháng 7 2021
a) Ta có: \(P=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)
\(=\left(\dfrac{2}{\sqrt{xy}}+\dfrac{1}{x}+\dfrac{1}{y}\right):\dfrac{x\sqrt{x}+y\sqrt{x}+x\sqrt{y}+y\sqrt{y}}{x\sqrt{xy}+y\sqrt{xy}}\)
\(=\left(\dfrac{x+2\sqrt{xy}+y}{xy}\right):\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
5 tháng 7 2021
a) Đk:\(x>0;y>0\)
\(P=\left[\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}.\sqrt{y}}.\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{x\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{xy}+y\sqrt{xy}}\)
\(=\left[\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\right]:\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)
\(=\dfrac{2\sqrt{xy}+x+y}{xy}:\dfrac{\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}.\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
b) \(xy=16\Leftrightarrow x=\dfrac{16}{y}\)
\(P=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=\dfrac{1}{\sqrt{\dfrac{16}{y}}}+\dfrac{1}{\sqrt{y}}=\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\)
Áp dụng AM-GM có:
\(\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\ge2\sqrt{\dfrac{\sqrt{y}}{4}.\dfrac{1}{\sqrt{y}}}=1\)
\(\Rightarrow P\ge1\)
Dấu "=" xảy ra khi \(y=4\Rightarrow x=4\)
Vậy x=y=4 thì P đạt GTNN là 1
25 tháng 10 2022
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
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2
C
20 tháng 12 2022
\(0,5\sqrt{100}-\dfrac{1}{4}\sqrt{16}+\left(-\dfrac{2}{3}\right)^2\\=\dfrac{1}{2}.10-\dfrac{1}{4}.4+\dfrac{4}{9}\\ =5-1+\dfrac{4}{9}\\ =4+\dfrac{4}{9}\\ =\dfrac{40}{9} \)
27 tháng 6 2021
a) Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=\dfrac{-\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{-\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)
\(=-2\sqrt{2}\)
b) Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
c) Ta có: \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
\(=\left(\dfrac{-\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}-2\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right)\)
\(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
\(=-\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=-1\)
27 tháng 6 2021
d) Ta có: \(\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
\(=\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
16 tháng 8 2023
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
\(=\dfrac{1}{2}\cdot10-\dfrac{1}{4}+1=6-\dfrac{1}{4}=\dfrac{23}{4}\)