Đáp án:
1/ a/ $x^2+6x+9$
b/ $x^2-5x+\dfrac{25}{4}$
c/ $\dfrac{9}{4}x^2+\dfrac{3}{2}x+\dfrac{1}{4}$
d/ $4x^2-12xy^2+9y^4$
2/ a/ $2x^2+2x+5$
b/ $2x+5$
3/ a/ $MIN_{A}=1$ khi $x=-3$
b/ $MIN_{B}=5$ khi $x=\dfrac{1}{2}$
Giải thích các bước giải:
1/ a/ $(x+3)^2=x^2+2.3.x+3^2=x^2+6x+9$
b/ `(x-\frac{5}{2})^2=x^2-2.\frac{5}{2}x+\frac{5^2}{2^2}=x^2-5x+\frac{25}{4}`
c/ `(\frac{3}{2}x+\frac{1}{2})^2`
`=(\frac{3}{2}x)^2+2.\frac{3}{2}x.\frac{1}{2}+\frac{1}{2^2}`
`=\frac{9}{4}x^2+\frac{3}{2}x+\frac{1}{4}`
d/ $(2x-3y^2)^2=(2x)^2-2.2x.3y^2+(3y^2)^2=4x^2-12xy^2+9y^4$
2/ a/ $(x-1)^2+(x+2)^2=x^2-2x+1+x^2+4x+4=2x^2+2x+5$
b/ $(2x-1)^2-(x-2)(x+2)-3x(x-2)$
$=4x^2-4x+1-(x^2-4)-3x^2+6x$
$=x^2+2x+1-x^2+4$
$=2x+5$
3/ a/ $A=x^2+6x+10=x^2+2.3x+9+1=(x+3)^2+1$
$\text{Vì $(x+3)^2 \geq 0$ nên $(x+3)^2+1 \geq 1$}$
$\text{Vậy GTNN của A là $1$ khi $x=-3$}$
b/ $B=6-4x+4x^2=4x^2-4x+1+5=(2x-1)^2+5$
$\text{Vì $(2x-1)^2 \geq 0$ nên $(2x-1)^2+5 \geq 5$}$
$\text{Vậy GTNN của B là $5$ khi $x=\dfrac{1}{2}$}$
