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\(A=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{\frac{9-2.3\sqrt{3}+3}{2}}=\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{\sqrt{2}}.\sqrt{\left(3-\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
=\(\sqrt{4}.\sqrt{3}+2\sqrt{9}.\sqrt{3}+3\sqrt{25}.\sqrt{3}-9\sqrt{16}.\sqrt{3}\)
=\(2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
=\(\left(2+6+15-36\right)\sqrt{3}\)
=\(-13\sqrt{3}\)
Lời giải:
Gọi biểu thức cần rút gọn là $P$
Xét tử số: $\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{3+2\sqrt{3.1}+1}-\sqrt{3}$
$=\sqrt{(\sqrt{3}+1)^2}-\sqrt{3}=|\sqrt{3}+1|-\sqrt{3}=1$
Xét mẫu số:
Ta dự đoán sẽ rút gọn được $\sqrt[3]{17\sqrt{5}-38}$
Đặt $17\sqrt{5}-38=(a+\sqrt{5})^3$ với $a$ nguyên.
$\Leftrightarrow 17\sqrt{5}-38=a^3+15a+\sqrt{5}(3a^2+5)$
$\Rightarrow 17=3a^2+5$ và $-38=a^3+15a$
$\Rightarrow a=-2$
Vậy $17\sqrt{5}-38=(-2+\sqrt{5})^3$
$\Rightarrow (\sqrt{5}+2)\sqrt[3]{17\sqrt{5}-38}=(\sqrt{5}+2)(-2+\sqrt{5})=1$
Vậy $P=\frac{1}{1}=1$
\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=5\sqrt{2}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{12}\)
\(=-15\sqrt{2}\)
Ta có : \(a^3=10+3\sqrt[3]{\left(5+\sqrt{52}\right)\left(5-\sqrt{52}\right)}\left(\sqrt[3]{5+\sqrt{52}}+\sqrt[3]{5-\sqrt{52}}\right)\)
\(=10+3\sqrt[3]{-27}.a=10-9a\)
\(\Rightarrow a^3+9a-10=0\Rightarrow\left(a-1\right)\left(a^2+a+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a^2+a+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\\left(a+\dfrac{1}{2}\right)^2+\dfrac{39}{4}>0\end{matrix}\right.\)
\(\Rightarrow a=1\) \(\Rightarrow f\left(a\right)=1+1+1^2+.....+1^{2015}=2016\)
\(\dfrac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\dfrac{2}{3}}+\dfrac{15}{\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{6}\left(5-\sqrt{6}\right)}{\sqrt{6}-5}-3.\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\)
\(=-2\sqrt{6}-\sqrt{3}.\sqrt{2}+\dfrac{15\left(\sqrt{6}+1\right)}{6-1}\)
\(=-2\sqrt{6}-\sqrt{6}+3\left(\sqrt{6}+1\right)\)
\(=3\).
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
@Nguyễn Thị Thu Sương :
\(\frac{\sqrt{3+\sqrt{15}}}{\sqrt{2}}=\sqrt{\frac{3+\sqrt{15}}{2}}\)
\(=\sqrt{\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{5-3}}\)
\(=\sqrt{\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}}\)
\(=\sqrt{\frac{\sqrt{3}}{\sqrt{5}-\sqrt{3}}}\)
a) \(\left(\sqrt{12}-\sqrt{27}+\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-3\sqrt{3}+\sqrt{3}\right):\sqrt{3}\)
\(=\sqrt{3}\left(2-3+1\right):\sqrt{3}\)
\(=0:\sqrt{3}=0\)
b) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
\(=\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
\(=\frac{5\sqrt{3}}{\sqrt{3}\cdot\sqrt{5}}+\frac{3\sqrt{5}}{\sqrt{3}\cdot\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{3}\)