\(\left[\left(a+32\right)-\left(a-32\right)\right]=64\Rightarrow\left(\sqrt{a+32}-\sqrt{a-32}\right)\left(\sqrt{a+32}+\sqrt{a-32}\right)=64\)
\(\Rightarrow4\left(\sqrt{a+32}+\sqrt{a-32}\right)=64\Rightarrow\sqrt{a+32}+\sqrt{a-32}=16\)
nếu\(\sqrt{a+32}-\sqrt{a-32}\) = 4 thì \(\sqrt{a+32}+\sqrt{a-32}\) = ?
Rút gọn biểu thức.
a) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
b) \(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}\)
A=\(\sqrt{12-\sqrt{80-32\sqrt{3}}}-\sqrt{12+\sqrt{80-32\sqrt{3}}}\) tính giá trị biểu thức
a,(\(\left(\sqrt{50}-\sqrt{32}+\sqrt{8}\right):\sqrt{2}\) b,\(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
Cho x,y,z>0 thỏa mãn \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\)\(1\).Tìm GTNN của:
\(A=\sqrt{\frac{x^2}{5x+32\sqrt{xy}+12y}}+\sqrt{\frac{y^2}{5y+32\sqrt{yz}+12z}}+\sqrt{\frac{z^2}{5z+32\sqrt{zx}+12x}}\)
\(A=2\sqrt{8}-3\sqrt{18}+4\sqrt{128}-5\sqrt{32}\)
\(A=\sqrt{50}-\dfrac{3}{2}\sqrt{32}-\dfrac{1}{3}\sqrt[]{72}+\sqrt{8}\)
Thực hiện phép tính (rút gọn biểu thức)
a)\(\left(\sqrt{3}-2\right)\sqrt{7+4\sqrt{3}}\)
b) \(\sqrt{6+\sqrt{32}}\) - \(\sqrt{11-\sqrt{72}}\)
c) \(\sqrt{21-4\sqrt{5}}\) + \(\sqrt{21+4\sqrt{5}}\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2=-3\\\sqrt[4]{x}+\sqrt{32-x}+6y=24\end{matrix}\right.\)
