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mik làm bài này
linh tinh
bn ơi
cho mik
xin 1 L-I-K-E
b,
d,
\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{\sqrt{5}-2}-\frac{2}{2+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}\)
\(=2\sqrt{5}+4-2\sqrt{5}+4\)
\(=8\)
\(\sqrt{25-2.5.\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
\(\sqrt{121+2.11.\sqrt{2}+2}=\sqrt{\left(11+\sqrt{2}\right)^2}=11+\sqrt{2}\)
\(\sqrt{\frac{9}{2}-2.\frac{3}{\sqrt{2}}.\frac{\sqrt{5}}{\sqrt{2}}+\frac{5}{2}}=\sqrt{\left(\frac{3}{\sqrt{2}}-\frac{\sqrt{5}}{\sqrt{2}}\right)^2}=\frac{3}{\sqrt{2}}-\frac{\sqrt{5}}{\sqrt{2}}=\frac{3\sqrt{2}-\sqrt{10}}{2}\)
\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{6\left(2+\sqrt{3}\right)}-4+\sqrt{2\left(2+\sqrt{3}\right)}}{2\sqrt{6}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}.\left(\frac{2\sqrt{6}}{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}\right)\)
\(=\frac{\sqrt{6\left(2+\sqrt{3}\right)}}{\left|\sqrt{3}+3\right|-4+\left|\sqrt{3}+1\right|}\)
\(=\frac{\left|\sqrt{3}+3\right|}{\sqrt{3}+3-4+\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+3}{2\sqrt{3}}\)
\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7-2\sqrt{10}}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}\right)^2+6\sqrt{2}+9}-\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{6}+\left(\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{10}+\left(\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+3-\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}+\sqrt{2}}\)
\(=\frac{3}{2\sqrt{2}+1}\)
\(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)
Vậy...
~ Chắc chắn đúng cậu nhé ~ Tiếc gì 1 tk cho tớ nào?
\(a,\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{2}\)
\(=\sqrt{3}+1-\sqrt{3}-\sqrt{2}+\sqrt{2}=1\)
\(b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{25}=5\)
\(A^3=14-3\sqrt[3]{\left(5\sqrt{2}+7\right)\left(5\sqrt{2}-7\right)}.A\)
\(\Leftrightarrow A^3=14-3A\)
\(\Leftrightarrow A^3+3A-14=0\)
\(\Leftrightarrow A=2\)
Ok thanh kiu bây bê (xl nếu hơi quá)