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a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
\(B=\frac{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}{3+\sqrt{5}}=3-\sqrt{5}\)
\(C=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)
\(=\frac{-2\sqrt{3}}{2}=-\sqrt{3}\)
\(D=\frac{2}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)
\(=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{6\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+2\right)-\left(3-\sqrt{3}\right)\)
\(=\sqrt{3}-1-\sqrt{3}-2-3+\sqrt{3}=\sqrt{3}-6\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
\(A=\left(\sqrt{8}-3\sqrt{2}+10\right)\left(\sqrt{2}-3\sqrt{0.4}\right)=\sqrt{16}-\frac{12\sqrt{5}}{5}+\sqrt{20}-6\sqrt{10}-6+\frac{18\sqrt{5}}{5}\)
\(A=-2+\frac{16\sqrt{5}}{5}-6\sqrt{10}\)
b)\(B=\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{6+2\sqrt{5}}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2}-\frac{\sqrt{5}-1}{2}=1\)
b) \(\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{5}-1}{2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}}{2}-\frac{\sqrt{5}-1}{2}\)
\(=\frac{\left(\sqrt{5}+1\right)-\sqrt{5}+1}{2}\)
\(=1\)
P/s: câu a) với câu c) vì ko có máy tính nên lười nháp, thông cảm, em tự làm đi