thank aj

a: A=x^2+4x+4+5

=(x+2)^2+5>=5

Dấu = xảy ra khi x=-2

b: =3/2(x^2+2/3x+2/3)

=3/2(x^2+2*x*1/3+1/9+5/9)

=3/2(x+1/3)^2+15/18>=15/18=5/6

Dấu = xảy ra khi x=-1/3

e: =x^2-2x+1+4

=(x-1)^2+4>=4

Dấu = xảy ra khi x=1

f: =2(x^2-3x)

=2(x^2-3x+9/4-9/4)

=2(x-3/2)^2-9/2>=-9/2

Dấu = xảy ra khi x=3/2

h: =-(x^2-4x-3)

=-(x^2-4x+4-7)

=-(x-2)^2+7<=7

Dấu = xảy ra khi x=2

Bài 2:

a: \(f\left(x\right)=x^3-9x^2+27x-27\)

\(=x^3-3x^2-6x^2+18x+9x-27\)

\(=x^2\left(x-3\right)-6x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\)

b: \(g\left(x\right)=x^2-6x+9\)

\(=x^2-2\cdot x\cdot3+3^2\)

\(=\left(x-3\right)^2\)

=>f(x)⋮g(x)

=>f(x) chia g(x) thì dư 0

\(\frac{f\left(x\right)}{k\left(x\right)}=\frac{x^3-9x^2+27x-27}{x^2-6x+10}\)

\(=\frac{x^3-6x^2+10x-3x^2+18x-30-x+3}{x^2-6x+10}=x-3+\frac{-x+3}{x^2-6x+10}\)

=>f(x) chia k(x) dư -x+3

a: AN+CN=AC

=>AN=20-15=5cm

Xét ΔABC có AM/AB=AN/AC

nên MN//BC

b: Xét ΔAMN và ΔNPC có

góc AMN=góc NPC(=góc B)

góc ANM=góc NCP)

=>ΔAMN đồng dạng với ΔNPC

\(\left(3-x\right)\left(x+1\right)-\left(2.x\right)\left(x+2\right)-3\)

\(=3x+3-x^2-x-2x^2-4x-3=-3x^2-2x\)

\(2x^2+2y^2-4xy+2x-2y+4\)

\(=2\left(x-y\right)^2+2\left(x-y\right)+4\)

\(=2\left[\left(x-y\right)^2+2\left(x-y\right)\frac{1}{2}+\frac{1}{4}\right]+\frac{7}{2}\)

\(=2\left(x-y+\frac{1}{2}\right)^2+\frac{7}{2}\)

\(\Rightarrow A\ge\frac{7}{2}\)

Dấu = bn tự tính nhé

1. 1² + 1²   = 1

2. (1+1)²  < 2

Làm J phải