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1:
a)10/20-15/20+16/20=-5/20+16/20
=11/20.
b)2/3+8/3:8/5=2/3+5/3
=7/3.
2:a)Ta có:
2x=1/4=3/4
2x=4/4=1
x=1:2
x=0,5
b)x:(2/12-1/12)=-3/8.
x:1/12=-3/8.
x=-3/8x1/12.
x=-1/32.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( x + 3 )3 : 3 - 1 = -10
( x + 3 )3 : 3 = -10 + 1
( x + 3 )3 = -9 * 3
x + 3 = \(\sqrt[3]{-27}\)
x = -3 - 3
x = -6
b) 3 | x - 1 | + 5 = 17
3 | x - 1 | = 17 - 5
| x - 1 | = 12 : 3
| x - 1 | = 4
( 1 ) x - 1 > 0 => x - 1 = 4 => x = 5
( 2 ) x - 1 < 0 => x - 1 = -4 => x = -3
Vậy S = { -3 ; 5 }
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\left(x-3\right)-3\left(x-1\right)=2x+6\)
\(\Leftrightarrow2x-6-3x+3=2x+6\)
\(\Leftrightarrow-3x=9\Leftrightarrow x=-3\)
Vậy x=-3
b) \(2\left(x+7\right)-4\left(x-2\right)=10\)
\(\Leftrightarrow2x+14-4x+8=10\)
\(\Leftrightarrow-2x=-12\Leftrightarrow x=-6\)
Vậy x=-6
c) \(10-\left(1-x\right)^2=6\Leftrightarrow\left(1-x\right)^2=4\)
\(\Leftrightarrow\left[\begin{matrix}1-x=-2\\1-x=2\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy x=3 hoặc x=-1
d)\(\left(x-3\right)^3-\left[\left(-2\right)^3+2^3\right]\left(1+3+5+7+...+2001\right)=-512\)
\(\Leftrightarrow\left(x-3\right)^3=-512\)
\(\Leftrightarrow x-3=-8\Leftrightarrow x=-5\)
Vậy x=-5
\(\frac{x+1}{x+3}=\frac{3}{4}\Rightarrow3.\left(x+3\right)=4\left(x+1\right)\)
\(\Rightarrow3x+9=4x+4\)
\(\Rightarrow3x-4x=4-9\)
\(\Rightarrow-x=-5\Rightarrow x=5\)
\(\frac{x+1}{x+3}=\frac{3}{4}\left(đkxđ:x\ne-3\right)\)
\(< =>\left(x+3\right).3=\left(x+1\right).4\)
\(< =>3x+9=4x+4\)
\(< =>4x-3x=9-4\)
\(< =>x=5\)