Tìm x a) 2x(x + 3) + 5(x + 3) = 0 b) x(x + 4) = 2(-x – 4) c) x^3 = x d) x^2 – 10x + 25 = 0 e) x^2 – 10x + 24 = 0 f) 2(x + 1)/ 3 – 1 = 3/2 – 1-2x / 4

a) 2x(x + 3) + 5(x + 3) = 0

⇔(x+3)(2x+5)=0

⇔\(\left[ \begin{array}{l}x+3=0\\2x+5=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=-3\\x=-5/2\end{array} \right.\) 

 Vậy x=-3; x=-5/2

b,x(x + 4) = 2(-x – 4)

⇔x(x+4)-2(-x-4)=0

⇔x(x+4)+2(x+4)=0

⇔(x+4)(x+2)=0

⇔\(\left[ \begin{array}{l}x+4=0\\x+2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=-4\\x=-2\end{array} \right.\) 

Vậy x=-4; x=-2

c,x³=x

⇔x³-x=0

⇔x(x²-1)=0

⇔x(x-1)(x+1)=0

⇔\(\left[ \begin{array}{l}x=0\\x-1=0\\x+1=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\) 

Vậy x=0;x=±1

d,x²-10x+25=0

⇔(x-5)²=0

⇔x-5=0

⇔x=5

Vậy x=5

e,x²-10x+24=0

⇔(x²-10x+25)-1=0

⇔(x-5)²-1²=0

⇔(x-5-1)(x-5+1)=0

⇔(x-6)(x-4)=0

⇔\(\left[ \begin{array}{l}x-6=0\\x-4=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=6\\x=4\end{array} \right.\) 

Vậy x=6;x=4

f,$\frac{2(x+1)}{3}$ -1=$\frac{3}{2}$ -$\frac{1-2x}{4}$ 

⇔$\frac{8(x+1)}{12}$ -12/12=$\frac{18}{12}$ -$\frac{3(1-2x)}{12}$ 

⇒8x+8-12=18-3+6x

⇔2x=19

⇔x=19/2

Vậy x=19/2

g,$\frac{3(x+1)}{10}$ = $\frac{2(x-1)}{3}$ + $\frac{4}{5}$

<=>$\frac{9(x+1)}{30}$= $\frac{20(x-1)}{30}$ +$\frac{24}{30}$

=>9x+9=20x-20+24

=>-11x=-5

<=>x=5/11

Vậy x=5/11

h,x²-x-2=0

⇔(x-2)(x+1)=0

⇔\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\) 

Vậy x=2;x=-1

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