a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

=>-4x=5

hay x=-5/4

b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

=>42x=41

hay x=41/42

`a)(10x+9)x-(5x-1)(2x+3)=8`

`<=>10x^2+9x-10x^2-15x+2x+3=8`

`<=>-4x=5`

`<=>x=-5/4`     Vậy `S={-5/4}`

`b)(3x-5)(7-5x)+(5x+2)(3x-2)-2=0`

`<=>21x-15x^2-35+25x+15x^2-10x+6x-4-2=0`

`<=>42x=41`

`<=>x=41/42`       Vậy `S={41/42}`

1/ (x+12)(x-3) = 0 nên x+12 = 0 hoặc x-3 = 0 => x = 0 - 12 = -12 hoặc x = 0 + 3 = 3.Vậy x \(\in\left\{-12;3\right\}\)

2/ (-x+5)(3-x) = 0 nên -x+5 = 0 hoặc 3-x = 0 => x = -(0-5) = -(-5) = 5 hoặc x = 3 - 0 = 3.Vậy x \(\in\left\{3;5\right\}\)

1)  - x + 5 = 0 hoặc 3 - x = 0 => x = 5 hoặc x = 3

2)   x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0 =. x = 0 hoặc x = - 2 hoặc x = 7

3)   x - 1 = 0 hoặc x + 2 = 0 hoặc - x - 3 = 0 => x = - hoặc x = - 2 hoặc x = - 3

1, => x + 12 = 0 => x = -12

x - 3 = 0 => x = 3 

=> x \(\in\) { -12; 3 } 

1; (\(x\) + 12)(\(x\) - 3) = 0

    \(\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)

    Vậy \(x\) \(\in\) { -12; 3}

1.

=3/5x(3/7+4/7)+2/5x(13/9-4/9)

=3/5x1+2/5x1

=3/5+2/5

=1

2.Xx(3/4+4/5)=7/10

Xx31/20=7/10

X         =7/10:31/20

X         =14/31

\(\frac{3}{5}\cdot\frac{3}{7}+\frac{3}{5}\cdot\frac{4}{7}+\frac{2}{5}\cdot\frac{13}{9}-\frac{2}{5}\cdot\frac{4}{9}\)

\(=\frac{3}{5}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{2}{5}\cdot\left(\frac{13}{9}-\frac{4}{9}\right)\)

\(=\frac{3}{5}\cdot1+\frac{2}{5}\cdot1\)\(=\frac{3}{5}+\frac{2}{5}=1\)

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\(\frac{3}{4}\cdot x+\frac{4}{5}\cdot x=\frac{7}{10}\)

\(\left(\frac{3}{4}+\frac{4}{5}\right)\cdot x=\frac{7}{10}\)

\(\frac{31}{20}\cdot x=\frac{7}{10}\)

\(x=\frac{7}{10}:\frac{31}{20}\)

\(x=\frac{14}{31}\)

3/ x(2 + x)(7 - x) = 0

=> x = 0 

hoặc 2 + x = 0 => x = -2

hoặc 7 - x = 0 => x = 7

Vậy x = 0, x = -2, x = 7

4/ (x - 1)(x + 2)(-x - 3) = 0

=> x - 1 = 0 => x = 1

hoặc x + 2 = 0 => x = -2

hoặc -x - 3 = 0 => -x = 3 => x = -3

Vậy x = 1 , x = -2 , x = -3

A) |x| = |-7|

|x| = 7

=>x=7  hoặc x=(-7)

Vậy x thuộc {7;-7}

B) |x+1|=2

=>x+1=2    hoặc x+1=(-2)

  x=2-1                x=(-2)-1

 x=1                    x=(-3)

Vậy x thuộc {1;-3}

C) |x+1|=3

=>x+1=3 hoặc x+1=(-3)

Vì x+1<0

nên x+1=(-3)

x=(-3)-1

x=(-4)

D) x +|-2| = 0

x+2=0

x=0-2

x=(-2)

E) 4.(3x – 4) – 2 = 18

4.(3x – 4) =18+2

4.(3x – 4) =20

3x-4=20 : 4

3x-4=5

3x=5+4

3x=9

x=9 : 3

x=3

a) \(\left|x\right|=\left|-7\right|\)

\(\Rightarrow\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

 Vậy ...

b) \(\left|x+1\right|=2\)

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

Vậy ...

d) \(x+\left|-2\right|=0\)

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)

Vậy ...

e) \(4\left(3x-4\right)-2=18\)

\(\Rightarrow4\left(3x-4\right)=20\)

\(\Rightarrow3x-4=5\)

\(\Rightarrow3x=9\Leftrightarrow x=3\)

Vậy ...

b)x^3 - 6x^2 +11x-6=0

<=>x^3 - x^2 - 5x^2 +5x + 6x - 6=0

<=>x^2(x - 1) - 5x(x - 1) +6(x - 1)=0

<=>(x-1).(x^2 - 5x + 6)=0

<=>(x - 1).(x^2 - 2x - 3x + 6)=0

<=>(x - 1).[(x(x-2)-3(x-2)]=0

<=>(x-1)(x-2)(x-3)=0

<=>x-1=0hoac x-2=0 hoac x-3=0

<=>x=1hoac x=2 hoac x=3

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