Bài 1:
`a)(3x+1)²`
`=(3x)²+2.3x.1+1²`
`=9x²+6x+1`
`b)(1/2+y)^2`
`=(1/2)^2+2. 1/2 .y+y²`
`=1/4+y+y²`
`c)(x+2y)²`
`=x²+2.x.2y+(2y)²`
`=x²+4xy+4y²`
`d)(x^2+2y)^2`
`=(x^2)^2+2.x².2y+(2y)²`
`=x^4+4x²y+4y²`
`e)(x+y+3)²`
`=x²+y²+3²+2.x.y+2.y.3+2.3.x`
`=x²+y²+9+2xy+6y+6x`
`f)(x^2+x+1)^2`
`=(x^2)^2+x²+1²+2.x².x+2.x.1+2.1.x²`
`=x^4+x²+1+2x³+2x+2x²`
`=x^4+2x³+3x²+2x+1`
Bài 2:
`a)(3x-2)^2`
`=(3x)^2-2.3x.2+2^2`
`=9x²-12x+4`
`b)(1-2y)^2`
`=1^2-2.1.2y+(2y)^2`
`=1-4y+4y^2`
`c)(2x-y)^2`
`=(2x)^2-2.2x.y+y^2`
`=4x^2-4xy+y^2`
`d)(x-y+2)^2`
`=x^2+y^2+2^2-2.x.y-2.y.2+2.2.x`
`=x^2+y^2+4-2xy-4y+4x`
Bài 3:
`a)x²+4x+4`
`=x²+2.x.2+2²`
`=(x+2)²`
`b)y²+8y+16`
`=y²+2.y.4+4²`
`=(y+4)²`
`c)4x²-4x+1`
`=(2x)²-2.2x.1+1²`
`=(2x-1)²`
`d)(y^2)/4-y+1`
`=(y/2)^2-2. y/2 .1+1^2`
`=(y/2-1)^2`
Bài 4:
`a)A=(x-2)²+1`
Ta có:`(x-2)²≥0∀x`
`⇒(x-2)²+1≥1∀x`
Vậy `A_(min)=1` khi `x-2=0⇔x=2`
`b)B=y²+14y+25`
`=y²+14y+49-24`
`=(y²+14y+49)-24`
`=(y²+2.y.7+7²)-24`
`=(y+7)²-24`
Ta có:`(y+7)²≥0∀y`
`⇒(y+7)²-24≥-24∀y`
Vậy `B_(min)=-24` khi `y+7=0⇔y=-7`
