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

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

\(A=\left(\frac{x-2}{x+2}-\frac{x+2}{2-x}-\frac{x^2-3x+6}{x^2-4}\right):\left(1-\frac{3}{x-2}\right)\)

\(=\left(\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-3x+6}{\left(x-2\right)\left(x+2\right)}\right)\)\(:\left(\frac{x-2}{x-2}-\frac{3}{x-2}\right)\)

\(=\frac{x^2-4x+4+x^2+4x+4-x^2+3x-6}{\left(x+2\right)\left(x-2\right)}:\frac{x-2-3}{x-2}\)

\(=\frac{x^2+3x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x-5}{x-2}\)

\(=\frac{x^2+x+2x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x-5}{x-2}\).

\(=\frac{x\left(x+1\right)+2\left(x+1\right)}{\left(x+2\right)\left(x-2\right)}.\frac{x-2}{x-5}\)

\(=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+2\right)\left(x-2\right)}.\frac{x-2}{x-5}\)

\(=\frac{x+1}{x-2}.\frac{x-2}{x-5}\)

\(=\frac{x+1}{x-5}\)

.

a) \(A=x^2-20x+101\)

\(=x^2-2.x.10+10^2+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu = xảy ra khi \(\left(x-10\right)^2=0\)

=> \(x-10=0\)

=> \(x=10\)

Vậy A min = 1 tại x = 10

b) \(B=4a^2+4a+2\)

\(=\left(2a\right)^2+2.2a.1+1^2+1\)

\(=\left(2a+1\right)^2+1\ge1\forall x\)

Dấu = xảy ra khi \(\left(2x+1\right)^2=0\)

=> \(2x+1=0\)

=> \(2x=-1\)

=> \(x=\frac{-1}{2}\)

Vậy B min = 1 tại \(x=\frac{1}{2}\)

c) Mình không biết làm mong bạn thông cảm

d)\(D=x^2+2y^2-2xy-4y+5\)

\(=x^2-2xy+y^2+y^2-2.y.2+2^2+1\)

\(=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\forall x\)

Dấu = xảy ra khi \(\hept{\begin{cases}\left(y-2\right)^2=0\\\left(x-y\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}y-2=0\\x-y=0\end{cases}}\Rightarrow\hept{\begin{cases}y=2\\x-2=0\end{cases}}\hept{\begin{cases}y=2\\x=2\end{cases}}\)

Vậy D min = 1 tại x = y = 2

(x+2)(x+3) - (x-2)(x+5) = 0

x² + 5x + 6 - x² - 3x + 10 = 0

2x + 16 = 0

2x = -16

x = -8

b. (9x² +12x + 4) + 6(3x + 2) + 9

= (3x+2)^2 + 2.3.(3x+2) + 3^2

= (3x+2+3)^2

= (3x+5)^2

( x - 5).(x + 5)² - (x - 5)²

= ( x - 5 ).(x² + 10x + 25 - x² + 10x - 25)

= ( x - 5 ).20

= 20x - 100