Question

a)(x-1)(x^2+5x-2)-x^3+1=0

b)5(x^2+3x)-9(3x+3)=x^2-36

Answer 1

a, Ta có : \(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\)

=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)

=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)

=> \(\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)

=> \(\left(x-1\right)\left(4x-3\right)=0\)

=> \(\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{1,\frac{3}{4}\right\}\)

b, Ta có : \(5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\)

=> \(5x^2+15x-27x-27=x^2-36\)

=> \(5x^2+15x-27x-27-x^2+36=0\)

=> \(4x^2-12x+9=0\)

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

=> \(x=\frac{3}{2}\)

Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2}\right\}\)

Answer 2

\(a.\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{1;\frac{3}{4}\right\}\)

\(b.5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27-x^2+36=0\\ \Leftrightarrow4x^2-12x+9=0\\ \Leftrightarrow\left(2x-3\right)^2=0\\ \Leftrightarrow x=\frac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\frac{3}{2}\right\}\)

Chúc bạn học tốt!!!!!!!!!!!