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\(A=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{4+2\sqrt{3}}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
=> \(A=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
=> \(A=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
=> \(A=4\sqrt{3}-8+6-4\sqrt{3}\)
=> \(A=-8+6=-2\)
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 \(B=2\)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
\(\left(3-\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right).\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3+\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)
\(=\left(3-\sqrt{5}\right)\left[\left(\sqrt{5}\right)^2-1\right]\)
\(=\left(3-\sqrt{5}\right)\left(5-1\right)\)
\(=4\left(3-\sqrt{5}\right)\)
\(=12-4\sqrt{5}\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
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/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)
\(=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{\sqrt{2}.\sqrt{2}.\sqrt{5}}{\sqrt{2}.\sqrt{3}}\\ =\dfrac{\sqrt{10}}{\sqrt{3}}=\dfrac{\sqrt{30}}{3}\)
\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\\ =\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =2\sqrt{5}:\sqrt{6}\\ =2\sqrt{5}.\dfrac{1}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}.\sqrt{6}}{6}\\ =\dfrac{1}{3}.\sqrt{30}\\ =\dfrac{\sqrt{30}}{3}\)