Đáp án `+` Giải thích các bước giải `!`
`1.`
`a)`
`4x^2y+8xy^2+4y^2`
`= 4y(x^2+2xy+y)`
`b)`
`4x^2-4xy+y^2-36`
`= (4x^2-4xy+y^2)-36`
`= (2x-y)^2-6^2`
`= (2x-y-6)(2x-y+6)`
`c)`
`x^4-2x^3+x^2-x^2y^2`
`= x^2(x^2-2x+1-y^2)`
`= x^2[(x^2-2x+1)-y^2]`
`= x^2[(x-1)^2-y^2]`
`= x^2(x-y-1)(x+y-1)`
`2.`
`a)`
`x(x+3)+x-3 = 0`
`<=> x^2+3x+x-3 = 0`
`<=> x^2+4x-3 = 0`
`<=> (x^2+4x+4)-7 = 0`
`<=> (x+2)^2 = 7`
`<=>` \(\left[ \begin{array}{l}x+2=\sqrt{7}\\x+2=-\sqrt{7}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\sqrt{7}-2\\x=-\sqrt{7}-2\end{array} \right.\)
Vậy `S= {\sqrt{7}-2; -\sqrt{7}-2}`
`b)`
`1+2x+x^2-4 = 0`
`<=> (x^2+2x+1)-4 = 0`
`<=> (x+1)^2-2^2 = 0`
`<=> (x+1-2)(x+1+2) = 0`
`<=> (x-1)(x+3) = 0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\x+3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
Vậy `S= {1; -3}`
