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5 tháng 9 2020
a) Ta có: \(A=2\sqrt{2+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12}+1}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{3+1-2\sqrt{3}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{3}-1}\)
\(\Leftrightarrow A=2\sqrt{\sqrt{3}+1}\)
\(\Leftrightarrow A\approx3,30578\)
b) Ta có: \(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(\Leftrightarrow B=\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)
\(\Leftrightarrow B=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(\Leftrightarrow B=\sqrt{2}.\left(4-2\right)\)
\(\Leftrightarrow B=2\sqrt{2}\)
\(\Leftrightarrow B\approx2,82843\)
NN
8 tháng 9 2020
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
8 tháng 9 2020
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)
WR
28 tháng 6 2019
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{4-3}=1\)
WR
28 tháng 6 2019
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}.\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
22 tháng 6 2018
\(B=\left(\sqrt{10}+\sqrt{6}\right).\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\) (vì\(\sqrt{5}-\sqrt{3}>0\))
\(=2\sqrt{2}\)
22 tháng 6 2018
\(A=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right).\sqrt{2}\)
\(=\sqrt{4}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{4}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{4}-\left|\sqrt{5}-1\right|\)
\(=\sqrt{4}-\sqrt{5}+1\) (vì \(\sqrt{5}-1>0\))
24 tháng 6 2019
\(a,\)\(\left(\sqrt{7}-\sqrt{5}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{7}-\sqrt{5}\right)\)
\(=\left[\left(\sqrt{2}-\sqrt{5}\right)+\sqrt{7}\right]\left[\left(\sqrt{2}-\sqrt{5}\right)-\sqrt{7}\right]\)
\(=\left(\sqrt{2}-\sqrt{5}\right)^2-\sqrt{7^2}\)
\(=2-2\sqrt{10}-5-7\)
\(=-10-2\sqrt{10}\)
24 tháng 6 2019
\(b,B=\sqrt{2}.\sqrt{2+\sqrt{3}}\)
\(\Rightarrow B^2=|2|.|2+\sqrt{3}|\)
\(=2.\left(2+\sqrt{3}\right)\)
\(=4+2\sqrt{3}\)
\(=3+2\sqrt{3}+1\)
\(=\sqrt{3}^2+2\sqrt{3}+\sqrt{1}^2\)
\(=\left(\sqrt{3}+\sqrt{1}\right)^2\)
\(\Rightarrow B=\sqrt{3}+\sqrt{1}=\sqrt{3}+1\)
14 tháng 6 2017
a, \(\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)+ \(\sqrt{\left(\sqrt{3}\right)^2-2\cdot\left(\sqrt{3}\right)\cdot\left(\sqrt{2}\right)+\left(\sqrt{2}\right)^2}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)+ \(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
= \(\sqrt{3}\)+ \(\sqrt{2}\)+ \(\sqrt{3}\)- \(\sqrt{2}\)= 2\(\sqrt{3}\)
14 tháng 6 2017
a) \(\sqrt{5+2\sqrt{6}+}\sqrt{5-2\sqrt{6}}\)
=\(\frac{\sqrt{10+4\sqrt{6}}}{\sqrt{2}}+\frac{\sqrt{10-4\sqrt{6}}}{\sqrt{2}}\)
=\(\frac{\sqrt{\left(\sqrt{6+2}\right)}^2}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{6-2}\right)}^2}{\sqrt{2}}\)
=\(\frac{\sqrt{6}+2}{\sqrt{2}}+\frac{\sqrt{6}-2}{\sqrt{2}}\)
=\(\frac{\sqrt{2\left(\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}+\frac{\sqrt{2\left(\sqrt{3}-\sqrt{2}\right)}}{\sqrt{2}}\)
=\(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)
=\(2\sqrt{3}\)
b )\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}-\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
=\(\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
=\(\frac{\sqrt{3}-1}{\sqrt{2}}-\frac{\sqrt{3}+1}{\sqrt{2}}\)
=\(\frac{-2}{\sqrt{2}}\)
=\(-\sqrt{2}\)
NN
1 tháng 11 2020
b) \(B=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
NN
1 tháng 11 2020
c) \(C=\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}+1\right|+\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}+1+2-\sqrt{2}=3\)
\(A^2=2+\sqrt{2+\sqrt{2+...}}\)
\(A^2=2+A\)
\(A^2-A-2=0\Rightarrow A^2-2A+A-2=0\)
\(\left(A-2\right)\left(A+1\right)=0\)
=> A = 2 hoặc A = -1 ( loại A > 0 )
Vậy A = 2