Question

Giải các phương trình sau bằng cách đưa về dạng ax + b = 0

a. x( x + 3 )² - 3x = ( x + 2 )³ + 1

b. ( x- 3 ) ( x + 4 ) - 2 ( 3x - 2) = (x - 4 )²

Answer 1

Câu a :

\(x\left(x+3\right)^2-3x=\left(x+2\right)^3+1\)

\(\Leftrightarrow x\left(x^2+6x+9\right)-3x=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+9x-3x-x^3-6x^2-12x-8-1=0\)

\(\Leftrightarrow-6x-9=0\) ( dạng ax+b=0)

\(\Rightarrow x=-\dfrac{3}{2}\)

Câu b :

\(\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8-x^2+8x-16=0\)

\(\Leftrightarrow3x-24=0\)

\(\Rightarrow x=8\)

Answer 2

hay wa