K = a² - 2ab + 5b² - 4b + 9

= (a² - 2ab + b²) + (4b² - 4b + 1) + 8

= (a - b)² + (2b - 1)² + 8

Do (a - b)² ≥ 0 với mọi a, b ∈ R

(2b - 1)² ≥ 0 với mọi b R

⇒ (a - b)² + (2b - 1)² ≥ 0 với mọi a, b ∈ R

⇒ (a - b)² + (2b - 1)² + 8 ≥ 8 với mọi a, b ∈ R

Vậy GTNN của K là 8 khi a = b = 1/2

m.n ơi giúp mk 1 hoặc 2 câu đc ko ạ mk cần gấp lắm mà mk ko bt cách lm

Bài 2 : 

a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)

Dấu ''='' xảy ra khi x = 2 

b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)

Dấu ''='' xảy ra khi x = -1 

 Bài 1 : 

a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)

c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(M=\left|3x-2\right|+\left|3x-6\right|=\left|3x-2\right|+\left|6-3x\right|\ge\left|3x-2+6-3x\right|=4\)

\(M_{min}=4\) khi \(\dfrac{3}{2}\le x\le2\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

không biết

\(A=4x^2-12x+9-\left(x^2+6x+5\right)+2\)

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

\(=3\left(x^2-6x+9\right)-21\)

\(=3\left(x-3\right)^2-21\ge-21\)

\(A_{min}=-21\) khi \(x=3\)

Ta có: \(A=\left(2x-3\right)^2-\left(x+1\right)\left(x+5\right)+2\)

\(=4x^2-12x+9-x^2-6x-5+2\)

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

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

\(=3\left(x^2-2\cdot x\cdot\dfrac{9}{2}+\dfrac{81}{4}-\dfrac{77}{4}\right)\)

\(=3\left(x-\dfrac{9}{2}\right)^2-\dfrac{231}{4}\ge-\dfrac{231}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{9}{2}\)

a) Ta có: \(A=x^2-3x+5\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

b: Ta có: \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)

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

\(=5x^2+5\ge5\forall x\)

Dấu '=' xảy ra khi x=0