\(A=\dfrac{5x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}=\dfrac{1}{x^2}-\dfrac{1}{x}+5=\left(\dfrac{1}{x^2}-\dfrac{1}{x}+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

\(A_{min}=\dfrac{19}{4}\) khi \(\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\)

\(2\left(x^2-\frac{3}{2}\right)^2+3\left(y-1\right)^2+\frac{4013}{2}\)

A=x^2-2.x.3+9+2000

=(x-3)^2+2000

Bước này bạn tự suy luận ra rồi kết luận là xong

\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2-6y+9\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge\frac{3}{4}\Rightarrow MinM=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=3\)\(P=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\Rightarrow MinP=4\Leftrightarrow x=1\)

a, Đặt tính chia ta được Q=2x+3,R=x²-4x+5

b,\(R=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\)

Vì (x-2)² >= 0 

=> R = (x-2)²+1 >= 1

Dấu "=" xảy ra <=> x-2=0 <=> x=2

Vậy GTNN của R =  1 khi x=2

\(M=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)

Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(M\ge\left|x-1+3-x\right|=\left|2\right|=2\)

Dấu " = " xảy ra khi \(x-1\ge0;3-x\ge0\)

\(\Rightarrow x\ge1;x\le3\)

\(\Rightarrow1\le x\le3\)

Vậy \(MIN_M=2\) khi \(1\le x\le3\)