TD
2
K
Khách
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B
1
12 tháng 3 2019
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{x+1}{2}\)
\(=\frac{1}{2}\left(2+3+4+...+x+1\right)\)
\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)
\(=\frac{1}{2}\cdot\frac{x\left(x+3\right)}{2}=\frac{x\left(x+3\right)}{4}\).
T
1
KK
1
MC
1
7 tháng 11 2021
\(=3:\left[\dfrac{4}{9}+\dfrac{1}{2}-\dfrac{4}{3}\right]-\dfrac{1}{7}\)
\(=3\cdot\dfrac{-18}{7}-\dfrac{1}{7}=\dfrac{-55}{7}\)
A
20 tháng 6 2018
\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)
\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)
\(\Rightarrow x=\frac{4}{93}\)
Vậy \(x=\frac{4}{93}\)
KK
1
3 tháng 11 2019
\(\frac{1}{2}\left(-2x+\frac{2}{3}\right)=2x+\frac{1}{2}\)
=> \(-x+\frac{1}{3}=2x+\frac{1}{2}\)
=> \(-x-2x=\frac{1}{2}-\frac{1}{3}\)
=> \(-3x=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\left(-3\right)\)
=> \(x=-\frac{1}{18}\)
PT
1
22 tháng 3 2023
\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=13/54
U
3
29 tháng 4 2020
1) x3 - 1 +x - x2
2) (a2 + 9)2 - 36a2
3) ( x2 + 1)2 - 4x2
4) ( 4a2 + 1/4)2 - 4a2
5) 81 - (x2 + 6x)2
6) 16a2 - (a2 + 4)2
7) 1/4 ( a + 1)2 - 4/9 ( a-2)2
8) ( x2 + xy)2 - (y2 + xy)2
9) 12a2b2 - 3( a2 +b2)2
10) 4x2y2 - ( x2 + y2 - a2)2
11) ( a+b+c)2 + ( a+b-c)2 - 4c2
12) x3 - 1 + 5x2 - 5 +3x - 3
13) ( x - y)2 + 4(x-y) + 4
14) x2 -2x( 3x+1) + (3x+1)2
15) x4 + 2x2(2x+1) + ( 2x+1)2
hok tốt
U
0
VM
13 tháng 11 2019
b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)
\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)
\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)
\(\Rightarrow x=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}.\)
c) \(5-\left|3x-1\right|=3\)
\(\Rightarrow\left|3x-1\right|=5-3\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)
d) \(\left(1-2x\right)^2=9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow1-2x=\pm3.\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;2\right\}.\)
Chúc bạn học tốt!
$=3+2\text{/}\left{1+3\text{/}\left[2-1\text{/}\left(3+2\text{/}\left(1-3\right)\right)\right]\right}$
$=3+2\text{/}\left{1+3\text{/}\left[2-1\text{/}\left(3+2\text{/}\left(-2\right)\right)\right]\right}$
$=3+2\text{/}\left{1+3\text{/}\left[2-1\text{/}\left(3+\left(-1\right)\right)\right]\right}$
$=3+2\text{/}\left{1+3\text{/}\left[2-1\text{/}2\right]\right}$
$=3+2\text{/}\left{1+3\text{/}\left[2-0.5\right]\right}$
$=3+2\text{/}\left{1+3\text{/}1.5\right}$
$=3+2\text{/}\left{1+2\right}$
$=3+2\text{/}3$
\(=\dfrac{3\cdot3+2}{3}\)
\(=\dfrac{11}{3}\)
$3+2/{1+3/[2-1/(3+2/(1-3))]}$
$3+2/{1+3/[2-1/(3+2/(-2))]}$
$3+2/{1+3/[2-1/(3+(-1))]}$
$3+2/{1+3/[2-1/2]}$
$3+2/{1+3/[2-0.5]}$
$3+2/{1+3/1.5}$
$3+2/{1+2}$
$3+2/3$
\(=\dfrac{3\cdot3+2}{3}\)
\(=\dfrac{11}{3}\)