Tìm x biết : a, -12( x - 5 ) + 7 ( 3 - x ) = 2016
b, 20 ( x - 4 ) - 5 ( x + 3 ) - 15x = 20
c, l 2x - 5 l = 13
d, ( x + 3 ) ( x - 7 ) < 0
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a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
-12(x-5)+7(3-x)=5
=> -12x+60+21-7x=5
=> -19x+81=5
=>-19x=5-81
=>-19x=-76
=>x=4
Vậy x=4
a) Ta có : x.710 = 712
=> x = 72
=> x = 49
b) 520 : x = 515
=> x = 55
=> x = 625
c) 7x + 1 = 50
=> 7x = 49
=> 7x = 72
=> x = 2
d) Sửa 7x + 1 = 23
=> 7x + 1 = 8
=> 7x = 7
=> x = 1
e) (x + 5)2 - 2 = 79
=> (x + 5)2 = 81
=> (x + 5)2 = 92
=> \(\orbr{\begin{cases}x+5=9\\x+5=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{4;-14\right\}\)
g) (7 - x)3 = 125
=> (7 - x)3 = 53
=> 7 - x = 5
=> x = 2
Vậy x =2