a) | \(\frac{1}{2}\)x| = 3 - 2x

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-\left(3-2x\right)\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x+2x=3\\\frac{1}{2}x=-3+2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=3\\\frac{1}{2}x-2x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3:\frac{5}{2}\\-\frac{3}{2}x=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-3:\left(-\frac{3}{2}\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=2\end{cases}}\)

b) |x - 1| = 3x + 2

\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-\left(3x+2\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x-1=-3x-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-2x=3\\x+3x=-2+1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{-2}\\4x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)

c) | 5x | = x - 12

\(\Rightarrow\orbr{\begin{cases}5x=x-12\\5x=-\left(x-12\right)\end{cases}}\Rightarrow\orbr{\begin{cases}5x-x=-12\\5x=-x+12\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=-12\\5x+x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\6x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

d) |7 - x| = 5x + 1

\(\Rightarrow\orbr{\begin{cases}7-x=5x+1\\7-x=-\left(5x+1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}7-1=5x+x\\7-x=-5x-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}6=6x\\7+1=-5x+x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\8=-4x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

e) |9 + x| = 2x

\(\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\Rightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}}\Rightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)

Ủng hộ mk nha !!! ^_^

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

a) \(\left|x+9\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=2x\\x+9=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)

b) \(\left|5x\right|-3x=2\Leftrightarrow\left|5x\right|=3x+2\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+2\\-5x=3x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)

c) \(\left|x+6\right|-9=2x\Leftrightarrow\left|x+6\right|=2x+9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=2x+9\\-x-6=2x+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

d) \(\left|2x-3\right|+x=21\Leftrightarrow\left|2x-3\right|=21-x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=x-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)

e) \(\left|2x+4\right|=-4x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=4x\\2x+4=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)

i) \(\left|3x-1\right|+2=x\Leftrightarrow\left|3x-1\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x-2\\3x-1=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)

g) \(\left|x+15\right|+1=3x\Leftrightarrow\left|x+15\right|=3x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-1\\x+15=1-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3,5\end{matrix}\right.\)

h) \(\left|2x-5\right|+x=2\Leftrightarrow\left|2x-5\right|=2-x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=3\end{matrix}\right.\)

a) |9+x|=2x

TH1: 9+x=2x

<=> 9=2x-x

<=> x=9

TH2: -9-x=2x

<=> -9=3x

<=> x=-3

b) |5x|-3x=2

TH1: 5x-3x=2

<=> 2x=2

<=> x=1

TH2: -5x-3x=2

<=> -8x=2

<=>x=-4

c) |x+6|-9=2x

TH1: x+6-9=2x

<=> -3=x

TH2: -x-6-9=2x

<=> -15=3x

<=>x=-5

d) |2x-3|+x=21

TH1: 2x-3+x=21

<=> 3x=24

<=> x=8

TH2: -2x+3+x=21

<=> -x=18

<=> x=-18

e,i,g,h tương tự

a) 4( 18 - 5x ) - 12( 3x - 16 ) = 15( 2x - 16 ) - 6( x + 14 )

<=> 72 - 20x - 36x + 192 = 30x - 240 - 6x - 84

<=> -20x - 36x - 30x + 6x = -240 - 84 - 72 - 192

<=> -80x = -588

<=> x = -588/-80 = 147/20

b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 6

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 6

<=> x² + 5x + 6 - x² - 3x + 10 = 6

<=> 2x + 16 = 6

<=> 2x = -10

<=> x = -5

c) -x( x + 3 ) + 2 = ( 4x + 1 )( x - 1 ) + 2x

<=> -x² - 3x + 2 = 4x² - 3x - 1 + 2x

<=> -x² - 3x - 4x² + 3x - 2x = -1 - 2

<=> -5x² - 2x = -3

<=> -5x² - 2x + 3 = 0

<=> -( 5x² + 2x - 3 ) = 0

<=> -( 5x² + 5x - 3x - 3 ) = 0

<=> -[ 5x( x + 1 ) - 3( x + 1 ) ] = 0

<=> -( x + 1 )( 5x - 3 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)

d) ( 2x + 3 )( x - 3 ) - ( x - 3 )( x + 1 ) = ( 2 - x )( 3x + 1 ) + 3 

<=> 2x² - 3x - 9 - ( x² - 2x - 3 ) = -3x² + 5x + 2 + 3

<=> 2x² - 3x - 9 - x² + 2x + 3 = -3x² + 5x + 2 + 3

<=> 2x² - 3x - x² + 2x + 3x² - 5x = 2 + 3 + 9 - 3

<=> 4x² - 6x = 11

<=> 4x² - 6x - 11 = 0

=> Vô nghiệm ( Lớp 8 chưa học nghiệm vô tỉ nên để vậy ) :))

vẫn làm được nha quỳnh !

\(4x^2-6x-11=0\)

\(< =>\left(4x^2-6x+\frac{9}{4}\right)-13\frac{1}{4}=0\)

\(< =>\left(2x-\frac{3}{2}\right)^2=\frac{53}{4}\)

\(< =>\orbr{\begin{cases}2x-\frac{3}{2}=\frac{\sqrt{53}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{53}}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}2x=\frac{3+\sqrt{53}}{2}\\2x=\frac{3-\sqrt{53}}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{3+\sqrt{53}}{4}\\x=\frac{3-\sqrt{53}}{4}\end{cases}}\)

:vc dcv ơi, lớp 8 chưa học số vô tỉ ớ nên cho PT đó vô nghiệm là đúng, cái j mà ko làm đc:)) haha ( pịt momx lại ko bị chửi mù chữ:)

a, \(4\left(18-5x\right)-12\left(3x-16\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Leftrightarrow72-20x-36x+192=30x-240-6x-84\)

\(\Leftrightarrow264-56x=24x-324\Leftrightarrow x=\frac{147}{20}\)

c, \(-x\left(x+3\right)+2=\left(4x+1\right)\left(x-1\right)+2x\)

\(\Leftrightarrow-x^2-3x+2=4x^2-4x+x-1+2x\)

\(\Leftrightarrow-x^2-3x+2=4x^2-x-1\)

\(\Leftrightarrow5x^2+2x-3=0\Leftrightarrow\left(x+1\right)\left(5x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)

\(\Leftrightarrow3x+2x-10=6-5x+1\)

\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm 

b, \(x^3-3x^2-x+3=0\)

\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)

Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }

c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow x+3+x^2-3x-2=0\)

\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn 

Vậy ... 

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2