1+1=2 haha
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27 tháng 5 2017
\(\dfrac{1}{3}x.\dfrac{2}{5}\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=0\\\dfrac{2}{5}\left(x-1\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy x = 0 hoặc x = 1
27 tháng 5 2017
\(\dfrac{1}{3}x.\dfrac{2}{5}\left(x-1\right)=0\\ =>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
A
3
5 tháng 11 2016
Có:
M = \(9+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+...+\frac{2}{8}+\frac{1}{9}\)
= \(9+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)-8\)
= \(1+10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}\right)\)
= \(10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\)
= 10N
=> \(\frac{m}{n}=10\)
20 tháng 4 2017
\(\left|1-3x\right|=\dfrac{1}{2}\)
\(\Rightarrow1-3x=\pm\dfrac{1}{2}\)
*)Xét \(1-3x=\dfrac{1}{2}\)
\(\Rightarrow3x=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{6}\)
*)Xét \(1-3x=-\dfrac{1}{2}\)
\(\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
24 tháng 12 2016
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}=\frac{1}{2}-\frac{\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\)
VC
0
Q
15 tháng 6 2017
\(A=\left(\dfrac{1+\sqrt{1-a}}{1-a+\sqrt{1-a}}+\dfrac{1+\sqrt{1+a}}{1+a+\sqrt{1+a}}\right)\cdot\dfrac{a^2-1}{2}+1\)
\(=\left(\dfrac{1+\sqrt{1-a}}{\sqrt{1-a}\cdot\left(\sqrt{1-a}+1\right)}+\dfrac{1+\sqrt{1+a}}{\sqrt{1-a}\cdot\left(\sqrt{1+a}+1\right)}\right)\cdot\dfrac{a^2-1}{2}+1\)
\(=\left(\dfrac{1}{\sqrt{1-a}}+\dfrac{1}{\sqrt{1+a}}\right)\cdot\dfrac{a^2-1}{2}+1\)
\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{\left(1-a\right)\cdot\left(1+a\right)}}\cdot\dfrac{a^2-1}{2}+1\)
\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1-a^2}}\cdot\dfrac{a^2-1}{2}+1\)
1 + 1 = 2
sao