Đáp án:
Giải thích các bước giải:
Bài `1` :
`2x^2-5x+3=0`
`<=>2x^2-4x+2-x+1=0`
`<=>2(x^2-2x+1)-(x-1)=0`
`<=>2(x-1)^2-(x-1)=0`
`<=>(x-1)[2(x-1)-1]=`
`<=>(x-1)(2x-2-1)=0`
`<=>(x-1)(2x-3)=0`
`<=>`\(\left[ \begin{array}{l}x-1=0\\2x-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy `x=1` hoặc `x=3/2`
Bài `2` :
`[xy(x+y)+xyz]+[yz(y+z)+xyz]+xz(x+z)`
`=(x^2y+xy^2+xyz)+(y^2z+yz^2+xyz)+xz(x+z)`
`=xy(x+y+z)+yz(y+z+x)+xz(x+z)`
`=(x+y+z)(xy+yz)+xz(x+z)`
`=y(x+y+z)(x+z)+xz(x+z)`
`=(x+z)[y(x+y+z)+xz]`
`=(x+z)(xy+y^2+yz+xz)`
`=(x+z)[y(x+y)+z(x+y)]`
`=(x+z)(x+y)(y+z)`
