Đáp án:
`↓↓↓`
Giải thích các bước giải:
Câu 1 :
`a) 4-x^2-2xy-y^2`
`=(x^2-2xy-y^2)-4`
`=-(x^2+2xy+y^2)-2²`
`=-(x+y)²-2²`
`=-(x+y-2)(x+y+2)`
`b) x^2-25`
`=x^2-5²`
`=(x-5)(x+5)`
`b) x^2-xy+x`
`=x(x-y+1)`
Câu 2 :
`a) (2x-3)^2-(x-1)^2=0`
`⇒(2x-3+x-1)(2x-3-x+1)=0`
`⇒(3x-4)(x-2)=0`
`⇒`\(\left[ \begin{array}{l}3x-4=0\\x-2=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=\frac{4}{3}\\x=2\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\frac{4}{3}\\x=2\end{array} \right.\)
`b) (x-2)^2-5x+10=0`
`⇒x²-4x+4-5x+10=0`
`⇒x^2-9x+14=0`
`⇒x^2+2x+7x+14=0`
`⇒x(x+2)+7(x+2)=0`
`⇒(x+7)(x+2)=0`
`⇒`\(\left[ \begin{array}{l}x+7=0\\x+2=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=-2\\x=-7\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=-2\\x=-7\end{array} \right.\)
