A=\(\frac{1\text{x}2+2\text{x}4+3\text{x}6+...+n\text{x}2n}{3\text{x}4+6\text{x}8+9\text{x}12+...+x}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{2\text{x}3+4\text{x}6+6\text{x}9+8\text{x}12}\)
\(\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{2\text{x}3+4\text{x}6+6\text{x}9+8\text{x}12}\)
\(=\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{\text{1x}2\text{x}3+2\text{x}4\text{x}3+3\text{x}6\text{x}3+4\text{x}8\text{x}3}\)
\(=\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{3\left(1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8\right)}\)
\(=\frac{1}{3}\)
Ta có: \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}=\frac{1.2}{2.3}+\frac{2.4}{4.6}+\frac{3.6}{6.9}+\frac{4.8}{8.12}.\)
\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{3}.4=\frac{4}{3}\)
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-7.2\right)}\)
\(\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.-9}=\frac{2.9}{-9}=-2\)
a: Đặt \(x^2-4=a\)
Pt sẽ là \(a=3\sqrt{xa}\)
\(\Rightarrow a^2=9xa\)
\(\Leftrightarrow a\left(a-9x\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)
hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)
d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)
Pt sẽ là 2a+b=ab+2
=>(b-2)(1-a)=0
=>b=2 và 1-a
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)