|x+1|+|x+2|+|x+3|+|x+3|+|x+4|=x-5
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a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
1: =>(x+2018)(6x-3)=0
=>x+2018=0 hoặc 6x-3=0
=>x=1/2 hoặc x=-2018
2: x(x-11)+3(11-x)=0
=>(x-11)(x-3)=0
=>x=11 hoặc x=3
4: =>(x+5)(2x-4)=0
=>2x-4=0 hoặc x+5=0
=>x=2 hoặc x=-5
3: =>(x-3)(x+2)=0
=>x=3 hoặc x=-2
Bài 1:
\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)
\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)
\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
\(x\left(x-11\right)+3\left(11-x\right)=0\)
\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)
Câu 3:
\(x\left(x-3\right)-2\left(3-x\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Câu 4:
\(2x\left(x+5\right)-4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
=> 3 ( x + 2 ) = 4 ( x - 5 )
=> 3x + 6 = 4x - 20
=> 3x - 4x = - 6 - 20
=> - 1x = - 26
=> x = 26
1) 3/x-5=-4/x+2
3x+6=-4x+20
7x . =14
x . =2
2) x+3/-4=-9/x+3
(x+3)^2=36
Ta có hai trường hợp:
*x+3=6=)x=3
*x+3=-6=)x=-9
a) ( x - 1 )( x2 + x + 1 ) + x( x + 2 )( 2 - x ) = 5
<=> x3 - 1 - x( x + 2 )( x - 2 ) = 5
<=> x3 - 1 - x( x2 - 4 ) = 5
<=> x3 - 1 - x3 + 4x = 5
<=> 4x - 1 = 5
<=> 4x = 6
<=> x = 6/4 = 3/2
b) 5x( x - 3 )2 - 5( x - 1 )3 + 15( x + 4 )( x - 4 ) = 5
<=> 5x( x2 - 6x + 9 ) - 5( x3 - 3x2 + 3x - 1 ) + 15( x2 - 16 ) = 5
<=> 5x3 - 30x2 + 45x - 5x3 + 15x2 - 15x + 5 + 15x2 - 240 = 5
<=> 30x - 235 = 5
<=> 30x = 240
<=> x = 8
a,\(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(< =>x^3-1+x\left(4-x^2\right)=5\)
\(< =>x^3-1+4x-x^3=5\)
\(< =>4x-1-5=0< =>4x-6=0< =>x=\frac{3}{2}\)
b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)
\(< =>5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)=5\)
\(< =>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)
\(< =>\left(5x^3-5x^3\right)+\left(15x^2+15x^2-30x^2\right)+\left(45x-15x\right)+5-240=5\)
\(< =>30x-240=5-5=0< =>x=\frac{24}{3}=8\)
\(|x+1|+|x+2|+|x+3|+|x+3|+|x+4|=x-5\left(1\right)\)
Ta có: \(|x+1|\ge0;\forall x\)
\(|x+2|\ge0;\forall x\)
\(|x+3|\ge0;\forall x\)
\(|x+4|\ge0;\forall x\)
\(\Rightarrow|x+1|+|x+2|+|x+3|+|x+3|+|x+4|\ge0;\forall x\)
Mà \(|x+1|+|x+2|+|x+3|+|x+3|+|x+4|=x-5\)
\(\Rightarrow x-5\ge0\)
\(\Rightarrow x\ge5\)
\(\Rightarrow|x+1|=x+1\)
\(|x+2|=x+2\)
..................................
\(|x+4|=x+4\)
Thay vào (1) ta được :
\(x+1+x+2+x+3+x+3+x+4=x-5\)
\(\Leftrightarrow5x+13=x-5\)
\(\Leftrightarrow4x=-18\)
\(\Leftrightarrow x=\frac{-9}{2}\)
Vậy \(x=\frac{-9}{2}\)