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

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

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^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

Đặt `3(x+2)-1/3(6-3x)=0`

`<=>3(x+2)-(2-x)=0`

`<=>3x+2+x-2=0`

`<=>4x=0`

`<=>x=0`

Vậy nghiệm của đa thức là 0

`3x(x-5)-(x+3x)=0`

`<=>3x(x-5)-4x=0`

`<=>x(3x-15-4)=0`

`<=>x(3x-19)=0`

`<=>[(x=0),(3x-19=0):}`

`<=>[(x=0),(x=19/3):}`

Vậy nghiệm đa thức là 0 và `19/3`.

a) Đặt \(3\left(x+2\right)-\dfrac{1}{3}\left(6-3x\right)=0\)

\(\Leftrightarrow3x+6-2+x=0\)

\(\Leftrightarrow4x=-4\)

hay x=-1

b) Đặt 3x(x-5)-(x+3x)=0

\(\Leftrightarrow3x^2-15x-4x=0\)

\(\Leftrightarrow x\left(3x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{3}\end{matrix}\right.\)

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=\(\dfrac{-11}{6}\)

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

 1/4 - 5/2 x |3x - 1/5|=2/3 x |3x - 1/5|- 2/3

Tương đương với 1/4+2/3 = 2/3 x l3x - 1/5l + 5/2 x l3x-1/5l

                           11/12    = l3x - 1/5l x (2/3 + 5/2)

                            11/12  = l3x -1/5 l x 19/6

        => l3x - 1/5l = 11/12 : 19/6 = 11/38

  Xét 2 trường hợp:

      + 3x - 1/5 = 11/38 => 3x = 11/38 + 1/5 = 93/190 => x = 93/190 : 3 = 31/190

      + 3x - 1/5 = -11/38 => 3x = -11/38 + 1/5 = -17/190 => x = -17/190 : 3 = -17/570