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bạn ơi hình như âu tính giá trị biểu thức N bị sai chỗ phân tích \(\sqrt{21-12\sqrt{3}}\)thì phải ,hình như phải bằng \(\left(2\sqrt{3}-3\right)^2\)
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{1}=1}\)
b) \(B=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{21-6\sqrt{12}}}=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{\left(3-2\sqrt{3}\right)^2}}}}=\sqrt{\sqrt{3}-\sqrt{2\sqrt{3}-2}}\)c)
\(C=\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}=\frac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{2+2\sqrt{5}}{\sqrt{2}}=\sqrt{2}+\sqrt{10}=\sqrt{2}\left(\sqrt{5}+1\right)\)
\(A=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(A=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(A=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)
\(x+y+z=2\sqrt{x-34}+4\sqrt{y-21}+6\sqrt{z-4}+45\)
ĐK: \(x\ge34;y\ge21;z\ge4\)
\(pt\Leftrightarrow x-34-2\sqrt{x-34}+1+y-21-4\sqrt{y-21}+4+z-4-6\sqrt{z-4}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-34}-1\right)^2+\left(\sqrt{y-21}-2\right)^2+\left(\sqrt{z-4}-3\right)^2=0\left(1\right)\)
Dễ Thấy: \(VT_{\left(1\right)}\ge0\) nên dấu "=" khi
\(\hept{\begin{cases}\sqrt{x-34}=1\\\sqrt{y-21}=2\\\sqrt{z-4}=3\end{cases}}\)
Giải tiếp rồi thay vào T
a) \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\sqrt{10}+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\sqrt{10}-1\)
b) \(\sqrt{21-6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}-\sqrt{3}\)
\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\frac{10+4\sqrt{6}}{2}}-\sqrt{\frac{10-4\sqrt{6}}{2}}\)
\(=\sqrt{\frac{6+2.2.\sqrt{6}+4}{2}}-\sqrt{\frac{6-2.2.\sqrt{6}+4}{2}}\)
\(=\frac{\sqrt{\left(\sqrt{6}+2\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{6}-2\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{6}+2\right|-\left|\sqrt{6}-2\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{6}+2-\sqrt{6}+2}{\sqrt{2}}\)
\(=\frac{4}{\sqrt{2}}\)
\(=2\sqrt{2}\)
a, \(\sqrt{11-2\sqrt{10}}\) = \(\sqrt{1-2\sqrt{10}+\sqrt{10}^2}\) = \(\sqrt{\left(1-\sqrt{10}\right)^2}\)
= \(\left|1-\sqrt{10}\right|\)
= \(\sqrt{10}-1\)
b, \(\sqrt{21-6\sqrt{6}}\) = \(\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\cdot\sqrt{2}\cdot\sqrt{3}+\sqrt{3}^2}\)
= \(\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
= \(\left|3\sqrt{2}-\sqrt{3}\right|\)
= \(3\sqrt{2}\) - \(\sqrt{3}\)
\(a.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.\sqrt{18}.\sqrt{3}+3}+\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}+\sqrt{18}-\sqrt{3}=2\sqrt{18}=6\sqrt{2}\)