a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1

<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5

<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3

<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5

<=> |x - 5/3|/6 = 9/5

<=> |x - 5/3| = 9/5.6

<=> |x - 5/3| = 54/5

<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5

       x = 54/5 + 5/3         x = -54/5 - 5/3

       x = 187/15              x = -137/15

b) 2.|3x + 1| = 1/3.|3x + 1| + 5

<=> 2.|3x + 1| - 1/3.|3x + 1| = 5

<=> 5/3.|3x + 1| = 5

<=> 5.|3x + 1| = 5.3

<=> 5.|3x + 1| = 15

<=> |3x + 1| = 15 : 5

<=> |3x + 1| = 3

       3x + 1 = 3 hoặc 3x + 1 = -3 

       3x = 3 - 1           3x = -3 - 1

       3x = 2                3x = -4

       x = 2/3               x = -4/3

=> x = 2/3 hoặc x = -4/3

c) làm tương tự câu a) mình hơi lời

Làm câu c) cho

\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)

\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)

\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)

Giải tiếp nha

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

1. Thu gọn biểu thức - Hoc24 làm rồi mà bạn?

1.

a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)

b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)

2.

a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)

b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

1,a,=x2−6x+8+3x2−15x=4x2−21x+8b,=9x2+12x+4−x2+9=8x2+12x+132,a,⇔x2+8x+16−x2+4=5⇔8x=−15⇔x=−158b,⇔9x2−6x+1−8x2−2x+12x+3−x2=5⇔4x=1⇔x=14

2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

k mk đi

ai k mk 

mk k lại

thanks

không ai trả lời 

a,\(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)

\(< =>6x-2-5x+15-18x+36=24\)

\(< =>-29x+49=24< =>29x=25< =>x=\frac{25}{29}\)

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

\(< =>2x^2+4x^2-4=6x^2+2x\)

\(< =>2x=-4< =>x=-\frac{4}{2}=-2\)

c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)

\(< =>10x-6x^2+6x^2-10x-3x+21=4\)

\(< =>-3x=4-21=-17< =>x=\frac{17}{3}\)

d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)

\(< =>5x^2+5x-4x^2-8x=1-x\)

\(< =>x^2-3x+x-1=0\)

\(< =>x^2-2x-1=0\)

\(< =>\left(x-1\right)^2=2\)

\(< =>\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)

Sai rồi bn:)

a, \(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)

\(\Leftrightarrow6x-2-5x+15-18x+36=24\)

\(\Leftrightarrow-17x+25=0\Leftrightarrow x=\frac{25}{17}\)

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

\(\Leftrightarrow2x^2+4x^2-4=6x^2+2x\)

\(\Leftrightarrow-4-2x=0\Leftrightarrow x=-2\)

c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)

\(\Leftrightarrow10x-6x^2+6x^2-10x-3x+21=4\)

\(\Leftrightarrow-3x+17=0\Leftrightarrow x=\frac{17}{3}\)

d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)

\(\Leftrightarrow5x^2+5x-4x^2-8x=1-x\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{cases}}}\)