Question

(3x-1)(x\(^2\)+2)=(3x-1)(7x-10)

Answer 1

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\\x=5\end{matrix}\right.\)

Vậy ...............

Answer 2

Mộc Lung Hoa bài của Thư đúng rồi đó nhé, mình làm thiếu

Answer 3

(3x-1)(x²+2)-(3x-1)(7x-10)=0

(3x-1)(x²+2-7x+10)=0

(3x-1)(x²-7x+12)=0

(3x-1)(x²-3x-4x+12)=0

(3x-1)x(x-3)-4(x-3)=0

(3x-1)(x-3)(x-4)=0

->x=1/3 ; x=3 ;x=4

Answer 4

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

<=>\(\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

<=>\(\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

<=>\(\left(3x-1\right)\left(x^2-7x+12\right)=0\)

<=>\(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

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

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

=> 3x-1=0<=> x=1/3

hoặc x-3=0<=>x=3

hoặc x-4=0<=>x=4

vậy........