\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)
\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)
\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)
\(A=3\sqrt{3}\)
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:
\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)
\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)
\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)
\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)
\(A=\left(18-10-5\right)\sqrt{3}\)
\(A=3\sqrt{3}\)
