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g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
\(A=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{4+2\sqrt{3}}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
=> \(A=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
=> \(A=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
=> \(A=4\sqrt{3}-8+6-4\sqrt{3}\)
=> \(A=-8+6=-2\)
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 \(B=2\)
\(\left(3-\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right).\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3+\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)
\(=\left(3-\sqrt{5}\right)\left[\left(\sqrt{5}\right)^2-1\right]\)
\(=\left(3-\sqrt{5}\right)\left(5-1\right)\)
\(=4\left(3-\sqrt{5}\right)\)
\(=12-4\sqrt{5}\)
\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
\(1a.\left(\sqrt{72}-3\sqrt{5}+2\sqrt{8}\right).\sqrt{2}+\sqrt{90}=\sqrt{144}-3\sqrt{10}+2.\sqrt{16}+3\sqrt{10}=12+8=20\) \(b.\left(\sqrt{\dfrac{1}{5}}-10\sqrt{\dfrac{27}{5}}+2\sqrt{5}\right):\sqrt{5}+6\sqrt{3}=\left(\sqrt{\dfrac{1}{5}}-30\sqrt{\dfrac{3}{5}}+2\sqrt{5}\right).\dfrac{1}{\sqrt{5}}+6\sqrt{3}=\dfrac{1}{5}-6\sqrt{3}+2+6\sqrt{3}=\dfrac{11}{5}\) \(2.\sqrt{\left(3-\sqrt{10}\right)^2}=\sqrt{10}-3\)
\(b.\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}=2+\sqrt{3}+2-\sqrt{3}=4\) \(c.\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}\)
a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)
b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)
d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)
C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
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
\(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(=2\sqrt{5}-6-2+\frac{6\sqrt{5}}{5}\)
\(=\frac{16\sqrt{5}-40}{5}\)
\(=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(=2\sqrt{5}-6-2+\frac{6\sqrt{5}}{5}=\frac{16\sqrt{5}-40}{5}\)