Question

1. Giair phương trình

a) (y + 1)(2 - y)+(y - 2)²+y² - 4 =0

b) x³+x² - 4x=4

Answer 1

a)\(\left(y+1\right)\left(2-y\right)+\left(y-2\right)^2+y^2-4\)\(=0\)

<=>\(2y-y^2+2-y+y^2-4y+4+y^2-4\)\(=0\)

<=>\(y^2-3y+2=0\)

<=>\(\left(y^2-2y\right)-\left(y-2\right)=0\)

<=>\(\left(y-2\right)\left(y-1\right)=0\)

=>\(\orbr{\begin{cases}y-2=0\\y-1=0\end{cases}}\)=>\(\orbr{\begin{cases}y=2\\y=1\end{cases}}\)

b)\(x^3+x^2-4x=4\)

<=>\(x^3+x^2-4x-4=0\)

<=>\(\left(x^3+x^2\right)-\left(4x+4\right)=0\)

<=>\(x^2\left(x+1\right)-4\left(x+1\right)=0\)

<=>\(\left(x+1\right)\left(x^2-4\right)=0\)

<=>\(\left(x+1\right)\left(x+2\right)\left(x-2\right)=0\)

=>   \(x+1=0\)

       \(x+2=0\)

       \(x-2=0\)

=> \(x=-1;-2;2\)