Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
= ( 7\(\sqrt{16\cdot3}\)+3\(\sqrt{9\cdot3}\)-3\(\sqrt{4\cdot3}\)) /\(\sqrt{3}\)
=(49\(\sqrt{3}\)+ 9\(\sqrt{3}\)-6 \(\sqrt{3}\)) /\(\sqrt{3}\)
=52
a/ Đề sai
b/ \(\sqrt{125}-4\sqrt{45}+3\sqrt{2}-\sqrt{80}=5\sqrt{5}-12\sqrt{5}+3\sqrt{2}-4\sqrt{5}\)
\(=-11\sqrt{5}+3\sqrt{2}\)
c/ \(2\sqrt{\frac{27}{4}}-\sqrt{\frac{48}{9}}-\frac{2}{5}\sqrt{\frac{75}{16}}=2.\frac{3\sqrt{3}}{2}-\frac{4\sqrt{3}}{3}-\frac{2}{5}.\frac{5\sqrt{3}}{4}\)
\(=3\sqrt{3}-\frac{4\sqrt{3}}{3}-\frac{\sqrt{3}}{2}=\sqrt{3}\left(3-\frac{4}{3}-\frac{1}{2}\right)=\frac{7\sqrt{3}}{6}\)
d/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)
\(A=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}=\sqrt{3^2-\left(\sqrt{5}\right)^2}=\sqrt{4}=2\)
\(B=\sqrt{150.27.96}=\sqrt{150}.\sqrt{27}.\sqrt{96}=5\sqrt{6}.3\sqrt{3}.4\sqrt{6}=360\sqrt{3}\)
\(C=\left(\sqrt{27}+\sqrt{48}\right)^2-\left(\sqrt{27}-\sqrt{48}\right)^2\)\(=\left[\left(\sqrt{27}+\sqrt{48}-\sqrt{27}+\sqrt{48}\right)\left(\sqrt{27}+\sqrt{48}+\sqrt{27}-\sqrt{48}\right)\right]\)
\(=2\sqrt{27}.2\sqrt{48}=2.3\sqrt{3}.2.4\sqrt{3}=144\)
\(D=\sqrt{137^2-88^2}-\sqrt{192^2-111^2}=\sqrt{\left(137+88\right)\left(137-88\right)}-\sqrt{\left(192+111\right)\left(192-111\right)}\)
\(=\sqrt{225.49}-\sqrt{303.81}=15.7-9.\sqrt{303}=9\left(\frac{35}{3}-\sqrt{303}\right)\)
\(E=\sqrt{\frac{225}{4}.\frac{81}{25}.\frac{49}{64}}=\frac{15}{2}.\frac{9}{5}.\frac{7}{8}=\frac{189}{16}\)
\(F=\sqrt{\frac{27}{25}}.\sqrt{\frac{49}{189}}.\sqrt{\frac{700}{99}}=\frac{3\sqrt{3}}{5}.\frac{7}{3\sqrt{21`}}.\frac{10\sqrt{7}}{3\sqrt{11}}=\frac{14}{3\sqrt{11}}\)
\(H=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{\frac{35}{5}}+\sqrt{\frac{21}{5}}\right]=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{7}+\sqrt{\frac{21}{5}}\right]\)
\(=\sqrt{105}.\left[\frac{\sqrt{75}-\sqrt{49}+\sqrt{147}}{\sqrt{35}}\right]=\sqrt{3}\left(12\sqrt{3}-7\right)=36-7\sqrt{3}\)
\(K=\sqrt{64.14.21.54}-\sqrt{35.45.12}=8.\sqrt{14}.\sqrt{21}.3\sqrt{6}-\sqrt{35}.3\sqrt{5}.2\sqrt{3}\)
\(=24.\sqrt{14.21.6}-6\sqrt{35.5.3}=24.42-30\sqrt{21}=30\left(\frac{168}{5}-\sqrt{21}\right)\)
\(A=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{4+2\sqrt{3}}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
=> \(A=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
=> \(A=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
=> \(A=4\sqrt{3}-8+6-4\sqrt{3}\)
=> \(A=-8+6=-2\)
VẬY \(A=-2\)
\(B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
=> \(B=\sqrt{8-2\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=> \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\)
=> \(B=\left(\sqrt{5}-\sqrt{3}\right)^2\left(4+\sqrt{15}\right)\)
=> \(B=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
=> \(B=32+8\sqrt{15}-8\sqrt{15}-30\)
=> \(B=2\)
VẬY \(B=2\)
\(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)=\sqrt{3}\cdot4\sqrt{3}=12\)
\(=\left(\sqrt{3}\cdot\sqrt{4}+\sqrt{9}\cdot\sqrt{3}-\sqrt{3}\right)\cdot\sqrt{3}\\ =3\cdot\left(2+3-1\right)=12\)