Bài 1:
a) $(2x+1)^2-4(x+2)^2=9$
`<=>`$4x^2+4x+1-4x^2-16x-16=9$
`<=>`$12x-15=9$
`<=>`$-12x=24$
`<=>`$x=-2$
b) $(3x-1)^2+2(x+3)^2+11(x+1)(1-x)=6$
`<=>`$9x^2-6x+1+2x^2+12x+18+11x-11x^2+11-11x=6$
`<=>`$6x+30=6$
`<=>`$6x=-24$
`<=>`$x=-4$
Bài 3:
a) $(\frac{x}{2}-2y)^2$
$=(\frac{x}{2})^2-2.\frac{x}{2}.2y+(2y)^2$
$=\frac{x^2}{4}-2.\frac{x}{2}.2y+4y^2$
$=\frac{x^2}{4}-x.2y+4y^2$
$=\frac{x^2}{4}-2xy+4y^2$
b) $(2a^2+b^2)^2$
$=(2a^2)^2+2.a^2.b^2+(b^2)^2$
$=4a^4+2.2a^2.b^2+b^4$
$=4a^4+4a^2b^2+b^4$
c) $(a^2+5)(a^2-5)$
$=(a^2)^2-5^2$
$=a^4-5^2$
$=a^4-25$
Bài 4:
a) $A=x^2+5x+7$
$A=x^2+5x+7=(x^2+2.\frac{5}{2}x+\frac{25}{4})+\frac{3}{4}=(x+\frac{5}{2})^2+\frac{3}{4}$
Mà $(x+\frac{5}{2})^2+\frac{3}{4}\geq0$`=>`$(x+\frac{5}{2})^2+\frac{3}{4}\geq\frac{3}{4}$
Vậy: $A_{min}=\frac{3}{4}$ $khi$ $x+\frac{5}{2}=0$`=>`$x=\frac{-5}{2}$
b) $B=6x-x^2-5$
$B=5x-x^2=-(x^2-5x)=-[(x-\frac{5}{2}-\frac{25}{4}]=-(x-\frac{5}{2})^2+\frac{25}{4}\leq\frac{25}{4}$
Vậy: $B_{max}=\frac{25}{4}$ $khi$ $x=\frac{5}{2}$
Bài 5: $40^2-39^2+38^2-...+2^2-1^2$
$=(40^2-39^2)+(38^2-37^2)+...+(2^2-1^2)$
$=(40-39)(40+39)+(38-37)(38+37)+...+(2-1)(2+1)$
$=40+39+38+37+...+2+1$
$=\frac{40.41}{2}=820$
Bài 2: Mình không biết làm, xin lỗi bạn nhé!
