a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)
a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)
Rút gọn \(A=\sqrt{8+2\cdot\sqrt{10+2\cdot\sqrt{5}}}+\sqrt{8-2\cdot\sqrt{10-2\cdot\sqrt{5}}}\)
TÍNH :
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}\cdot\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(B=\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(C=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(D=\left[4+\sqrt{15}\right]\left[\sqrt{10}-\sqrt{6}\right]\cdot\sqrt{4-\sqrt{15}}\)
\(E=\left[3-\sqrt{5}\right]\cdot\sqrt{3+\sqrt{5}}\text{ }+\left[3+\sqrt{5}\right]\cdot\sqrt{3-\sqrt{5}}\)
1. Rút gọn biểu thức:
a) \(\sqrt{27\cdot48\cdot\left(1-a\right)^2}\)với a>1
b) \(\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}\) với a>b
c) \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}\)với \(a\ge0\)
d) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}\)với a>0
e) \(\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
M=\(\frac{1+ab}{a+b}-\frac{1-ab}{a-b}\)
với a=\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
b=\(\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
Tính M
Chứng minh: \(\frac{1}{2\cdot\sqrt{1}}+\frac{1}{3\cdot\sqrt{2}}+\frac{1}{4\cdot\sqrt{3}}+...+\frac{1}{2012\cdot\sqrt{2011}}+\frac{1}{2013\cdot\sqrt{2012}}\)\(< 2\)
Chứng minh: A=\(\frac{1}{3\cdot\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\cdot\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\cdot\left(\sqrt{48}+\sqrt{49}\right)}\)\(< \frac{1}{2}\)
Tính
a) \(\left(2-\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
b) \(\left(2\sqrt{3}-\sqrt{5}\right)\cdot\left(2\sqrt{3}+\sqrt{5}\right)\)
1. Tính:
a. \(\text{[}\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\sqrt{\frac{1}{ab}}\text{]}\cdot\sqrt{ab}\)
b.\(\text{[}-\frac{am}{b}\cdot\sqrt{\frac{n}{m}}-\frac{ab}{n}\cdot\sqrt{mn}+\frac{a^2}{b^2}\cdot\sqrt{\frac{m}{n}}\text{]}\cdot\text{[}a^2b^2\cdot\sqrt{\frac{n}{m}}\text{]}\)
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))
Tính:
a)\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}}\cdot\sqrt{18}-\sqrt{128}}}\)
b)\(\sqrt{6+2\cdot\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)