a, = \(3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)

= \(12\sqrt{2}-2\sqrt{3}\)

b, = 5 - 7 - 8 + 2

= - 8

= 5+(-7) - 2.4 - \(\dfrac{1}{3}\).6

= - 12

Có

\(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-\sqrt{75}=3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}=12\sqrt{2}-2\sqrt{3}\)

\(\sqrt[3]{125}+\sqrt[3]{-343}-2\sqrt[3]{64}+\frac{1}{3}\sqrt[3]{126}=5-7-8+\frac{1}{3}\sqrt[3]{126}=\frac{1}{3}\sqrt[3]{126}-10\)

a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)

b) 

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)

c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)

a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=3\sqrt{3}\)

vậy \(A=3\sqrt{3}\)

\(B=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)  \(ĐKXĐ:x>0;x\ne1\)

\(B=\left[1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(B=\left[1+\sqrt{x}\right]\left[1-\sqrt{x}\right]\)

\(B=1-x\)

vậy \(B=1-x\)

\(C=\sqrt[3]{64}-\sqrt[3]{-125}+\sqrt[3]{216}\)

\(C=\sqrt[3]{4^3}-\sqrt[3]{\left(-5\right)^3}+\sqrt[3]{6^3}\)

\(C=4+5+6\)

\(C=15\)

vậy \(C=15\)

Cho mk giải câu a:

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=\left(18-10-5\right)\sqrt{3}\)

\(A=3\sqrt{3}\)

a) \(\sqrt{28}+\sqrt{125}-3\sqrt{343}-\frac{3}{8}\sqrt{396}=2\sqrt{7}+5\sqrt{5}-21\sqrt{7}-\frac{9\sqrt{11}}{4}\)

\(=-19\sqrt{7}+5\sqrt{5}-\frac{9\sqrt{11}}{4}\)

b) \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\left(3-2\sqrt{2}\right)+2\sqrt{2}+3=6\)

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{2}\)

\(b,\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(7-5\right)\)

\(=-2\)