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a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
Lời giải:
a)
\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{3+1-2\sqrt{3}}\)
\(=\sqrt{(3+1+2\sqrt{3})+2+(2\sqrt{6}+2\sqrt{2})}-\sqrt{(\sqrt{3}-\sqrt{1})^2}\)
\(=\sqrt{(\sqrt{3}+1)^2+2\sqrt{2}(\sqrt{3}+1)+2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{(\sqrt{3}+1+\sqrt{2})^2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{3}+1+\sqrt{2}-(\sqrt{3}-1)=2+\sqrt{2}\)
b)
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right)(\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)\)
\(=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
Lời giải:
a)
\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{3+1-2\sqrt{3}}\)
\(=\sqrt{(3+1+2\sqrt{3})+2+(2\sqrt{6}+2\sqrt{2})}-\sqrt{(\sqrt{3}-\sqrt{1})^2}\)
\(=\sqrt{(\sqrt{3}+1)^2+2\sqrt{2}(\sqrt{3}+1)+2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{(\sqrt{3}+1+\sqrt{2})^2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{3}+1+\sqrt{2}-(\sqrt{3}-1)=2+\sqrt{2}\)
b)
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right)(\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)\)
\(=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)\)
\(=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)\)
\(=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}\)
\(=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139\)
Ta có: \(-14,5\sqrt{2}\simeq-20,506\)
\(VT\ne VP\)
Đẳng thức không xảy ra
a) \(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
Biến đổi vế trái :
VT = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\left|\sqrt{3}+1\right|}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{3}+3}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)\left(\sqrt{3}-3\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{\sqrt{2}\left(6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3\right)}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}=VP\left(đpcm\right)\)
b) \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Biến đổi vế trái :
VT = \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\sqrt{5+\sqrt{21}}\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\)
\(=\sqrt{2}\sqrt{5+\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{25-21}=\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{4}=\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)2\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)2=\left(7-3\right)2=4.2=8=VP\left(đpcm\right)\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2