B=\(4x^2-4x+1+x^2+4x+4=5x^2+5\)

                                                  \(=5\left(x^2+1\right)\)

vì\(x^2+1\ge1\forall x\)

\(\Leftrightarrow B\ge5\forall x\)

dấu'=' xảy ra \(\Leftrightarrow x^2+1=0\Leftrightarrow x=0\)

vậy B đạt GTNN =5 khi x=0

Bài 2: 

a) Ta có: \(A=x^2-3x+5\)

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

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

Ta có: \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)

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

Dấu '=' xảy ra khi \(x-\dfrac{3}{2}=0\)

hay \(x=\dfrac{3}{2}\)

Vậy: Giá trị nhỏ nhất của biểu thức \(A=x^2-3x+5\) là \(\dfrac{11}{4}\) khi \(x=\dfrac{3}{2}\)

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

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

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

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

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

\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)

\(A=\left|2021-x\right|+\dfrac{1}{2}\left|4040-2x\right|\)

\(A=\left|2021-x\right|+\left|2020-x\right|\)

\(A=\left|2021-x\right|+\left|x-2020\right|\ge\left|2021-x+x-2020\right|=1\)

\(A_{min}=1\) khi \(2020\le x\le2021\)

a)2x^2-4xy+4y^2+2x+5=x^2-4xy+4y^2+x^2+2x+1+4=(x-2y)^2+(x+1)^2+4>=4(dấu = tự tìm nhé)

b)x(1-x)(x-3)(4-x)=x(x-1)(x-3)(x-4)

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

Đặt x^2-4x=t(t>=-4) bt viết lại t(t+3)=t^2+3t>=-9/4

Dấu= xảy ra khi t=-3/2 >>>tìm x

\(A=\dfrac{1}{2}+\left|2x-1\right|\ge\dfrac{1}{2}\forall x\)

\(minA=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{\left|x\right|+2007}{2008}\ge\dfrac{0+2007}{2008}=\dfrac{2007}{2008}\)

\(minB=\dfrac{2007}{2008}\Leftrightarrow x=0\)

\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)

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

\(B=5x^2+5\)

Ta có: \(5x^2\ge0\forall x\)

\(\Rightarrow5x^2+5\ge5\forall x\)

\(B=5\Leftrightarrow5x^2=0\Leftrightarrow x=0\)

Vậy \(B_{min}=5\Leftrightarrow x=0\)

Tham khảo nhé~

B\(=\left(2x-1\right)^2+\left(x+2\right)^2\)

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

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

Dấu "=" xảy ra khi x=0

Vậy ...