Bài 3: Những hằng đẳng thức đáng nhớ

\(F=x^2+6x-5=x^2+6x+9-14=\left(x+3\right)^2-14\)

\(G=x^2-8x-5=x^2-8x+16-21=\left(x-4\right)^2-21\)

\(H=x^2-10x+31=x^2-10x+25+6=\left(x-5\right)^2+6\)

\(K=x^2-12x+9=x^2-12x+36-27\)

\(=\left(x-6\right)^2-27\)

\(M=x^2-14x+20=x^2-14x+49-29\)

\(=\left(x-7\right)^2-29\)

\(N=x^2-16x+25=x^2-16x+64-39\)

\(=\left(x-8\right)^2-39\)

\(P=9x^2+12x+5\)

\(=9x^2+12x+4+1=\left(3x+2\right)^2+1\)

\(S=9x^2+6x+3=9x^2+6x+1+2=\left(3x+1\right)^2+2\)

Ta có: 

\(a^3+2c=3ab\)

\(\Rightarrow\left(x+y\right)^3+2\left(x^3+y^3\right)=3\cdot\left(x+y\right)\left(x^2+y^2\right)\)

\(\Rightarrow\left(x^3+3x^2y+3xy^2+y^3\right)+2x^3+2y^3=3\left(x^3+xy^2+x^2y+y^3\right)\)

\(\Rightarrow x^3+3x^2y+3xy^2+y^3+2x^3+2y^3=3x^3+3xy^2+3xy^2+3y^3\)

\(\Rightarrow3x^3+3x^2y+3xy^2+3y^3=3x^3+3x^2y+3xy^2+3y^3\)

\(\Rightarrow\left(3x^3-3x^3\right)+\left(3x^2y-3x^2y\right)+\left(3xy^2-3xy^2\right)+\left(3y^3-3y^3\right)=0\)

\(\Rightarrow0=0\left(dpcm\right)\)

x³ - 27 = x³ - 3³

= (x - 3)(x² + 3x + 3²)

= (x - 3)(x² + 3x + 9)

(x + y)² = x² + 2xy + y²

\(\left(A+B\right)^2=A^2+2\cdot A\cdot B+B^2\)

\(\left(A+B\right)^2=A^2+2AB+B^2\)

Lời giải:

$x^2+4x-8y-2xy+2y^2+2024$

$=(x^2-2xy+y^2)+4x-8y+y^2+2024$

$=(x-y)^2+4(x-y)+y^2-4y+2024$

$=(x-y)^2+4(x-y)+4+(y^2-4y+4)+2016$

$=(x-y+2)^2+(y-2)^2+2016\geq 2016$

Vậy GTNN của biểu thức là 2016. Giá trị này đạt tại $x-y+2=y-2=0$

$\Leftrightarrow y=2; x=0$

a: \(A=2x^2-5x-2x\left(x+1\right)\)

\(=2x^2-5x-2x^2-2x\)

=-7x

b: \(B=\left(x+2y\right)\left(x-2y\right)-4y\left(x-y\right)\)

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

\(=x^2-4xy\)

c: \(C=\left(2x-8\right)\left(x^2+4x+16\right)-2x\left(x^2-2\right)\)

\(=2\left(x-4\right)\left(x^2+4x+16\right)-2x\left(x^2-2\right)\)

\(=2\left(x^3-64\right)-2x^3+4x\)

\(=2x^3-128-2x^3+4x=4x-128\)

a) \((a+y)^3=a^3+3a^2y+3ay^2+y^3\)

b) \((x-b)^3=x^3-3x^2b+3xb^2-b^3\)

a: \(\left(2n+1\right)^2-\left(2n-1\right)^2\)

\(=\left(2n+1-2n+1\right)\left(2n+1+2n-1\right)\)

\(=2\cdot4n=8n⋮8\)

b: \(\left(8n+4\right)^2-\left(2n+1\right)^2\)

\(=16\left(2n+1\right)^2-\left(2n+1\right)^2\)

\(=15\left(2n+1\right)^2⋮15\)

a: \(\left(a+b\right)^2-\left(a-b\right)^2\)

\(=a^2+2ab+b^2-\left(a^2-2ab+b^2\right)\)

\(=a^2+2ab+b^2-a^2+2ab-b^2\)

=4ab

b: \(\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)

\(=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)

c: \(2\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2+\left(a-b\right)^2\)

\(=\left(a+b\right)^2+2\cdot\left(a+b\right)\cdot\left(a-b\right)+\left(a-b\right)^2\)

\(=\left(a+b+a-b\right)^2=\left(2a\right)^2=4a^2\)