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
g) 3(x + 1) / 10 = 2(x – 1) / 3 + 4/5
h) x^2 – x – 2 = 0
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
g) 3(x + 1) / 10 = 2(x – 1) / 3 + 4/5
h) x^2 – x – 2 = 0
Đáp án:
Giải thích các bước giải:
a) 2x(x + 3) + 5(x + 3) = 0
⇒(x+3)(2x+5)=0
⇒x+3=0 hay 2x+5=0
⇒x=-3 hay x=$\frac{-5}{2}$
b) x(x + 4) = 2(-x – 4)
⇒ x(x + 4) =-2(x+4)
⇒x(x + 4) +2(x+4)=0
⇒(x+4)(x+2)=0
⇒x+4=0 hay x+2=0
⇒x=-4 hay x=-2
c)x³=x
⇒x³-x=0
⇒x(x²-1)=0
⇒x=0 hay x²-1=0
⇒x=0 hay (x-1)(x+1)=0
⇒x=0 hay x-1=0 ;x+1=0
⇒x=0 hay x=±1
d)x²-10x+25=0
⇒(x-5)²=0
⇒x-5=0
⇒x=5
e)x²-10x+24=0
⇒x²-6x-4x+24=0
⇒x(x-6)-4(x-6)=0
⇒(x-6)(x-4)=0
⇒x-6=0 hay x-4=0
⇒x=6 hay x=4
f)$\frac{2(x+1)}{3}$ -1=$\frac{3}{2}$ -$\frac{1-2x}{4}$
⇒$\frac{2x+2-3}{3}$ =$\frac{6-1+2x}{4}$
⇒$\frac{2x-1}{3}$ =$\frac{5+2x}{4}$
⇒$\frac{2x-1}{3}$-$\frac{5+2x}{4}$=0
⇒$\frac{(2x-1).4}{12}$ -$\frac{(5+2x)}{12}$ =0
⇒8x-4-15-6x=0
⇒2x-19=0
⇒x=$\frac{19}{2}$
g)$\frac{3(x+1)}{10}$ =$\frac{2(x-1)}{3}$ +$\frac{4}{5}$
⇒$\frac{3(x+1)}{10}$ -$\frac{2(x+1)}{3}$ -$\frac{4}{5}$ =0
⇒(3x+3).3-(2x-2).10-24=0
⇒9x+9-20x+20-24=0
⇒-11x+5=0
⇒x=$\frac{5}{11}$
h)x²-x-2=0
⇒x²-2x+x-2=0
⇒x(x-2)+(x-2)=0
⇒(x-2)(x+1)=0
⇒x-2=0 hay x+1=0
⇒x=2 hay x=-1
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