(a) Điều kiện : \(x\ne-1.\)

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

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

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

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

\(=x^2-x.\)

Vậy : Với mọi \(x\ne-1\) thì \(P=x^2-x.\)

(b) Ta có : \(P=x^2-x\)

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

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

Vậy : \(MinP=-\dfrac{1}{4}.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=\dfrac{1}{2}.\)

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Lời giải:
$A=x^2+2x+2xy+2y^2+4y+2021$

$=(x^2+2xy+y^2)+2x+y^2+4y+2021$

$=(x+y)^2+2(x+y)+1+(y^2+2y+1)+2019$

$=(x+y+1)^2+(y+1)^2+2019\geq 2019$

Vậy $A_{\min}=2019$ khi $x+y+1=y+1=0$

$\Leftrightarrow (x,y)=(0,-1)$

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

\(P=x^2-3x+\dfrac{1}{2x}+\dfrac{7}{4}+\dfrac{1}{4}\)

\(P=\dfrac{4x^3-12x^2+7x+2}{4x}+\dfrac{1}{4}=\dfrac{\left(x-2\right)\left(4x^2-4x-1\right)}{4x}+\dfrac{1}{4}\)

\(P=\dfrac{\left(x-2\right)\left[4x\left(x-2\right)+\dfrac{1}{2}\left(x-2\right)+\dfrac{7x}{2}\right]}{4x}+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(P_{min}=\dfrac{1}{4}\) khi \(x=2\)

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

\(P=x^2-4x+4+x+\dfrac{4}{x}-\dfrac{7}{2x}-2\)

\(P=\left(x-2\right)^2+x+\dfrac{4}{x}-\dfrac{7}{2x}-2\)

Áp dụng bđt cosi và bđt x \(\ge\)2

Ta có: P \(\ge0+2\sqrt{x\cdot\dfrac{4}{x}}-\dfrac{7}{2.2}-2=\dfrac{1}{4}\)

Dấu "=" xảy ra <=> x = 2

Vậy MinP = 1/4 <=> x = 2

a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)

\(B=x\left(2x-1\right)=2x^2-x=2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{1}{8}=2\left(x-\dfrac{1}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(minB=-\dfrac{1}{8}\Leftrightarrow x=\dfrac{1}{4}\)

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

\(minC=-\dfrac{4}{3}\Leftrightarrow x=-\dfrac{2}{3}\)

`B=x(2x-1)`

`=2x(x-1/2)`

`=2(x^2-1/2x)`

`=2(x^2-1/2x+1/16)-1/8`

`=2(x-1/4)^2-1/8>=-1/8`

Dấu "=" xảy ra khi `x=1/4`

`C=x(3x+4)`

`=3x(x+4/3)`

`=3(x^2+4/3x)`

`=3(x^2+4/3x+4/9)-4/3`

`=3(x+2/3)^2-4/3>=-4/3`

Dấu "=" xảy ra khi `x=-2/3`