Question

Giải phương trình
a) 2(9x^2 + 6x + 1) = (3x+1)(x-2)
b) 12/1-9x^2 = 1-3x/1+3x - 1+3x/1-3x

Answer 1

a/2(9x²+6x+1)=(3x+1)(x-2)

⇔2(3x+1)²= (3x+1)(x-2)

⇔ 2(3x+1)² :(3x+1)=x-2

⇔ 2(3x+1)=x-2

⇔6x+2-x+2=0

⇔5x+4=0

⇔5x=-4

⇔x=\(\frac{-4}{5}\)

b/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

⇔\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

⇔12=(1-3x)²-(1+3x)²

⇔-(1-3x-1-3x)(1-3x+1+3x)=--12

⇔-(-6x.2)=-12

⇔12x=-12

⇔x=-1

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