![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow-2x+3\cdot\left\{12-2\left[3x-20-2x-4x\right]+1\right\}=45\)
\(\Leftrightarrow-2x+3\cdot\left\{12-2\cdot\left(-3x-20\right)+1\right\}=45\)
=>-2x+3(11+6x+40)=45
=>-2x+153+18x=45
=>16x=45-153=-108
hay x=-6,75
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :-2x + 3x . { 12-2.[ 3.x-(20+2.x) - 4.x ] + 1 } = 45
<=>-2.x +3.x. [ 12-2 .( 3.x - 20 - 2.x - 4.x )+1 ) = 45
<=> -2.x + 3.x . [ 12- 2 ( -3.x -20 ) + 1 ] = 45
<=> -2.x + 3.x . ( 12 + 6.x + 40 + 1 ) = 45
<=>-2.x + 3.x . ( 53 + 6.x ) = 45
<=> -2.x + 159.x + 18.x2 = 45
<=> 157 .x + 18.x2 = 45
<=> 157.x + 18.x2 - 45 = 0
<=> 18.x2 - 5.x + 162.x - 45 = 0
<=> x. ( 18.x-5 ) + 9 .( 18.x-5) = 0
<=> ( 18.x-5 ) . ( x + 9 ) = 0
<=> \(\orbr{\begin{cases}18.x-5=0\\x+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=-9\end{cases}}\)
Vậy x = { \(\frac{5}{18};-9\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\frac{1}{2}=\frac{x}{3}\)
\(\Rightarrow3x+\frac{3}{2}=x\)
\(\Rightarrow x-3x=\frac{3}{2}\)
\(\Rightarrow-2x=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}:-2\)
\(\Rightarrow x=\frac{3}{2}.-\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)
Vậy \(x=-\frac{3}{4}\)
\(x+\frac{1}{2}=\frac{x}{3}\)
\(x=\frac{x}{3}-\frac{1}{2}\)
\(\frac{6x}{6}=\frac{2x}{6}-\frac{3}{6}\)
\(\Rightarrow6x=2x-3\)
\(6x-2x=3\)
\(4x=3\)
\(x=\frac{3}{4}\)
\(\frac{1-2x}{9}=\frac{1}{1-2x}\)
\(\Rightarrow\left(1-2x\right).\left(1-2x\right)=1.9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
\(\frac{4}{5}=\frac{12}{x}=\frac{y}{20}=\frac{8.\left(y-x\right)}{z}\)
\(\Rightarrow\frac{4}{5}=\frac{12}{x}\)
\(4x=60\)
\(x=15\)
\(\frac{4}{5}=\frac{y}{20}\)
\(y5=80\)
\(y=16\)
\(\frac{4}{5}=\frac{8\left(y-x\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{8.\left(80-15\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{520}{z}\)
\(\Leftrightarrow4z=2600\)
\(z=650\)
Vậy ... ( bạn tự kết luận nhé )
chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(\frac{x+1}{2}=\frac{x}{3}\)
<=> 3(x+1)=2x
<=> 3x+3-2x=0
<=> x+3=0
<=> x=-3
2, \(\frac{1-2x}{9}=\frac{1}{1-2x}\left(x\ne\frac{1}{2}\right)\)
<=> (1-2x)2=9
<=> \(\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\left(tmđk\right)}\)
Vậy x={-1;2}
![](https://rs.olm.vn/images/avt/0.png?1311)
x = 1
x = \(\frac{1}{2}\)= 0,5
x = 0