Help!!!!!!!!!!!!!!!! Tìm x

1, 3(5x-1)-x(x-2)+$x^{2}$-13x=7

⇔ 15x-3-$x^{2}$ +2x +$x^{2}$-13x-7=0

⇔ 4x -10 = 0

⇔ x = $\frac{10}{4}$ = $\frac{5}{2}$ 

Vậy S={ $\frac{5}{2}$ }

2, 4(x+2)-7(2x-1)+9(3x-4)=30

⇔ 4x +8 -14x+7 +27x -36 -30 =0

⇔ 17x -51 = 0

⇔ x = 3

Vậy S={3}

3, 2(5x-8)-3(4x-5)=4(3x-4)+11

⇔ 10x -16 -12x +15 -12x + 16 -11 = 0

⇔ -14x + 4 =0

⇔ -14x = -4

⇔ x = $\frac{4}{14}$ = $\frac{2}{7}$ 

Vậy S={ $\frac{2}{7}$ }

4, 3x(x-2) -3 ($x^{2}$ +1) = $x^{2}$ + 1 - x(x-2)

⇔ $3x^{2}$  - 6x - $3x^{2}$ -3 - $x^{2}$ - 1 + $x^{2}$  - 2x = 0

⇔ -8x -4 = 0

⇔ x = $\frac{-1}{2}$ 

Vậy S= { $\frac{-1}{2}$ }

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

⇔ 15x +25 -8x +12 -5x - 6x - 36 = 0

⇔ -4x + 1 =0 

⇔ x = $\frac{1}{4}$ 

Vậy S={ $\frac{1}{4}$ }

6, (7x+7)+3x(2x-1)-2x(3x+15) = -4

⇔ 7x +7 + $6x^{2}$ - 3x - $6x^{2}$ -30x +4 =0 

⇔ -26x + 11 =0

⇔ x = $\frac{11}{26}$ 

Vậy S={ $\frac{11}{26}$ }

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