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5 tháng 1 2020

\(\sqrt{\left(\sqrt{5}-3\right)^2}-\sqrt{\frac{1}{5}}\)

\(\Leftrightarrow|\sqrt{5}-3|-\sqrt{\frac{1}{5}}\)

\(\Leftrightarrow3-\sqrt{5}-\sqrt{\frac{1}{5}}\)

\(\Leftrightarrow\frac{15}{5}-\frac{5\sqrt{5}}{5}-\frac{\sqrt{5}}{5}\)

\(\Leftrightarrow\frac{15-5\sqrt{5}-\sqrt{5}}{5}\)

\(\Leftrightarrow\frac{15-6\sqrt{5}}{5}.\)

a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=4-2\sqrt{2}\)

b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)

 

24 tháng 8 2023

Sai rồi đấy ạ  câu P

29 tháng 6 2023

1)

\(\dfrac{5}{\sqrt{5}}=\dfrac{5\sqrt{5}}{5}\sqrt{5}\)

\(\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}}=\sqrt{\dfrac{3}{2}}\)

\(\dfrac{5}{\sqrt{7}}=\dfrac{5\sqrt{7}}{\sqrt{49}}=\left(\dfrac{5}{7}\right)\sqrt{7}\)

 

29 tháng 6 2023

lười :v

21 tháng 7 2021

Giải giúp em vs ạ

21 tháng 7 2021

bạn viết rõ đề ra nhé 

a, \(\left|3x+1\right|-x-5=0\Leftrightarrow\left|3x+1\right|=x+5\)ĐK : \(x\ge-5\)

TH1 : \(3x+1=x+5\Leftrightarrow x=2\)( tm )

TH2 : \(3x+1=-x-5\Leftrightarrow x=-\dfrac{3}{2}\)( tm )

4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)

\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)

\(=\dfrac{-\sqrt{5}}{2}\)

a: \(=\left(1+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{-\left(\sqrt{5}-1\right)}\right)\left(\sqrt{5}+1\right)\)

=(1-căn 5)(1+căn 5)

=1-5=-4

b: \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)

29 tháng 8 2021

11.

\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)

\(=\dfrac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\dfrac{\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+5+10\sqrt{5}}{20}+\dfrac{25+5-10\sqrt{5}}{20}\)

\(=3\)

29 tháng 8 2021

12.

\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{2+\sqrt{3}}{4-3}\)

\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)

\(=\sqrt{2}\)

1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

\(=3+2\sqrt{2}+\sqrt{5}-2\)

\(=2\sqrt{2}+\sqrt{5}+1\)

2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)

\(=3+2\sqrt{2}-3+2\sqrt{2}\)

\(=4\sqrt{2}\)

2: \(\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{2}-1}-\dfrac{1}{\sqrt{5}-2}\)

\(=\left(2\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}+1\right)-\sqrt{5}-2\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}-\sqrt{5}-2\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-2\sqrt{5}-2\)

3: \(=2\cdot3\sqrt{3}-6\cdot\dfrac{1}{\sqrt{3}}+2-\sqrt{3}-3\sqrt{3}\)

\(=6\sqrt{3}-2\sqrt{3}+2-4\sqrt{3}=2\)

30 tháng 8 2023

1) \(\dfrac{3+\sqrt{3}}{\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)

\(=\dfrac{\sqrt{5}\cdot\left(3+\sqrt{3}\right)}{\sqrt{5}\cdot\sqrt{5}}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\dfrac{3\sqrt{5}+\sqrt{15}}{5}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}\)

\(=\dfrac{3\sqrt{5}+\sqrt{15}}{5}-\left(\sqrt{3}-1\right)\)

\(=\dfrac{3\sqrt{5}+\sqrt{15}-5\sqrt{3}+5}{5}\)

2) \(\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{2}-1}-\dfrac{1}{\sqrt{5}-2}\)

\(=\dfrac{\left(2\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}}{2-1}-\dfrac{\sqrt{5}+2}{5-4}\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}-\left(\sqrt{5}+2\right)\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-2\sqrt{5}-2\)

3) \(2\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{1}{2}+\dfrac{\sqrt{3}-9}{\sqrt{3}}\)

\(=2\cdot3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{1}{2}+\dfrac{\sqrt{3}\left(1-3\sqrt{3}\right)}{\sqrt{3}}\)

\(=6\sqrt{3}-\dfrac{\sqrt{3}\cdot2\sqrt{3}}{\sqrt{3}}+\dfrac{1}{2}+1-3\sqrt{3}\)

\(=6\sqrt{3}-2\sqrt{3}+\dfrac{1}{2}+1-3\sqrt{3}\)

\(=\dfrac{1}{2}+1+\sqrt{3}\)

\(=\dfrac{3}{2}+\sqrt{3}\)

28 tháng 8 2023

\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{2\sqrt{2}}{2+\sqrt{2}}-\dfrac{4\sqrt{2}+4}{2+\sqrt{2}}\)

\(=\sqrt{3}+\sqrt{3}+\dfrac{-2\sqrt{2}-4}{2+\sqrt{2}}\)

\(=2\sqrt{3}+\dfrac{-2\left(2+\sqrt{2}\right)}{2+\sqrt{2}}\)

\(=2\sqrt{3}-2\)

\(------\)

\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)

\(=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\dfrac{5\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\dfrac{5\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right)}\)

\(=\dfrac{4\sqrt{5}-4}{5-1}+\dfrac{5\sqrt{5}-10}{5-4}+\dfrac{5\sqrt{5}-15}{5-9}\)

\(=5\sqrt{5}-10+\left(\dfrac{4\sqrt{5}-4}{4}+\dfrac{5\sqrt{5}-15}{-4}\right)\)

\(=\dfrac{4\cdot\left(5\sqrt{5}-10\right)}{4}+\left(\dfrac{4\sqrt{5}-4}{4}-\dfrac{5\sqrt{5}-15}{4}\right)\)

\(=\dfrac{20\sqrt{5}-40}{4}+\dfrac{-\sqrt{5}+11}{4}\)

\(=\dfrac{19\sqrt{5}-29}{4}\)

#Ayumu