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a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{5}}\)
\(\frac{6-\sqrt{6}}{\sqrt{6}-1}+\frac{6+\sqrt{6}}{\sqrt{6}}\)\(=\frac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\frac{6}{\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{6}}\)\(=\sqrt{6}+\frac{6}{\sqrt{6}}+1\)\(=\sqrt{6}\left(1+\frac{\sqrt{6}}{\sqrt{6}}\right)+1\)\(=\sqrt{6}\left(1+1\right)+1\)\(=\sqrt{6}.2+1\)
\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)\(=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-I\sqrt{20}-3I}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)\(=\sqrt{\sqrt{5}-I\sqrt{5}-1I}\)\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)\(=\sqrt{1}=1\)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
a) Ta có: \(\sqrt{16-6\sqrt{7}}+\sqrt{7}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+7}+\sqrt{7}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}\)
\(=\left|3-\sqrt{7}\right|+\sqrt{7}\)
\(=3-\sqrt{7}+\sqrt{7}\)
\(=3\)
b) Ta có: \(\sqrt{\left|12\sqrt{5}-29\right|}+\sqrt{12\sqrt{5}+29}\)
\(=\sqrt{\left(\sqrt{29-12\sqrt{5}}+\sqrt{12\sqrt{5}+29}\right)^2}\)
\(=\sqrt{29-12\sqrt{5}+2\sqrt{\left(29-12\sqrt{5}\right)\left(12\sqrt{5}+29\right)}+12\sqrt{5}+29}\)
\(=\sqrt{58+2\sqrt{121}}\)
\(=\sqrt{58+2.11}\)
\(=\sqrt{80}=4\sqrt{5}\)
\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)ta có:
\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3B\sqrt[3]{25-52}\)
\(=10-9B\)
Giải PT: \(B^3+9B-10=0\Leftrightarrow B^3-1+9B-9=0\)\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+1\right)+9\left(B-1\right)=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+10\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}B-1=0\\B^2+2B+1+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+1\right)^2=-9\left(L\right)\end{cases}}}\)
Vậy \(B=1\)
À chết mình làm nhầm, phải là \(\left(B-1\right)\left(B^2+B+1\right)\) nha, \(\left(B-1\right)\left(B^2+B+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}B=1\\B^2+2.\frac{1}{2}B+\frac{1}{4}-\frac{1}{4}+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2=-\frac{7}{4}\left(L\right)\end{cases}}\)
\(\sqrt{4\left(1-x\right)^2}-6=0\)
<=> \(\left|2\left(1-x\right)\right|=6\)
TH1: x \(\ge\)1 Khi đó pt trở thành:
\(2\left(x-1\right)=6\)
<=> x - 1 = 3
<=> x = 4 (tm)
TH2: x < 1, khi đó pt trở thành:
2(1 - x) = 6
<=> 1 - x = 3
<=> x = -2(tm)
vậy S= {4; -2}
Trả lời:
\(\sqrt{4\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow2.\left|1-x\right|=6\)
\(\Leftrightarrow\left|1-x\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}1-x=3\\1-x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)
Vậy \(x=\left\{-2,4\right\}\)
\(\sqrt{4x^2+4x+1}=x+2\)\(\left(x\ge-2\right)\)
\(\Leftrightarrow4x^2+4x+1=\left(x+2\right)^2\)
\(\Leftrightarrow4x^2+4x+1=x^2+4x+4\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\x=-1\left(TM\right)\end{cases}}\)
Vậy \(x=\left\{1,-1\right\}\)
\(\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{20-12\sqrt{5}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-2\sqrt{5}+3}}\)
a,
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=a-2\sqrt{ab}+b=\left(\sqrt{a}-\sqrt{b}\right)^2\)
b,
A=\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+2\sqrt{12}}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-1-\sqrt{12}}}}{\sqrt{6}+\sqrt{2}}\)\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)
B=
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\)\(\sqrt{\sqrt{5}-\sqrt{3-\left|2\sqrt{5}-3\right|}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)
\(=\sqrt{1}=1\)
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