tìm x:
a) 3/4+1/4:(x-1)= -2
b) /x-6/-5=0
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18 tháng 1 2024
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
NL
0
NL
1
8 tháng 10 2021
\(a,\Leftrightarrow x^2-x-x^2+4x-4=2\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(6-x\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-6+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\\ c,\Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
2 tháng 7 2021
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
15 tháng 3 2022
a: =>x+5>0
hay x>-5
b: =>2x+1<0
hay x<-1/2
c: =>(x-1)(x-4)>0
=>x>4 hoặc x<1
24 tháng 1 2021
a) Ta có: \(\left|4-5x\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-5x+4=24\\-5x+4=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=20\\-5x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)
b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
Vậy: \(x\in\left\{-8;6\right\}\)
MN
24 tháng 1 2021
a) \(\left|4-5x\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
b) \(\left(8+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
L
19 tháng 5 2022
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
28 tháng 8 2021
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
NT
0
a) 3/4 + 1/4 : (x-1)= -2
1/4 : (x-1) = -2- 3/4
1/4 : (x-1) = -8/4-3/4
1/4 : (x-1) = -11/4
x-1 = 1/4 : (-11/4)
x-1 = 1/4 . (-4/11)
x- 1= -1/11
x= -1/11+1
x= -1/11+ 11/11
x=10/11
Vậy x=10/11
a) 3/4 + 1/4 : ( x-1) = -2
1/4 : ( x-1) = -2 -3/4
1/4 : ( x-1) = -11/4
x-1 = 1/4 : -11/4
x-1 = 1/4 . -4/11
x-1 = -11/6
x - 1 = -11/6 + 1
x = 5/16
Vậy x = 5/16
b) |x-6| -5 = 0
=> x-6 = 0 + 5
x-6 = 5
x = 5+ 6
x = 11
Vậy x = 11