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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: \(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(\Leftrightarrow B=\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)
\(\Leftrightarrow B=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(\Leftrightarrow B=\sqrt{2}.\left(4-2\right)\)
\(\Leftrightarrow B=2\sqrt{2}\)
\(\Leftrightarrow B\approx2,82843\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{4-3}=1\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}.\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
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 \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
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. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)
\(B=\left(\sqrt{10}+\sqrt{6}\right).\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\) (vì\(\sqrt{5}-\sqrt{3}>0\))
\(=2\sqrt{2}\)
\(A=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right).\sqrt{2}\)
\(=\sqrt{4}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{4}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{4}-\left|\sqrt{5}-1\right|\)
\(=\sqrt{4}-\sqrt{5}+1\) (vì \(\sqrt{5}-1>0\))
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)