\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)

\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

\(P=x^2+4xy+4y^2-4xy-4y^2+2x+3\)

\(=x^2+2x+3\)

Lời giải:

$A=(x^2+4y^2+4xy)+x^2+5-8x-12y$

$=(x+2y)^2-6(x+2y)+x^2+5-2x$

$=(x+2y)^2-6(x+2y)+9+(x^2-2x+1)-5$

$=(x+2y-3)^2+(x-1)^2-5\geq 0+0-5=-5$

Vậy $A_{\min}=-5$. Giá trị này đạt được khi $x+2y-3=x-1=0$

$\Leftrightarrow x=1; y=1$

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

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

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

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

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

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

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

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

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

\(A_{\left(x,y\right)}=x^2+4y^2+1-4xy+2x-4y\)

Đặt 2y=z

\(A_{\left(x,z\right)}=x^2+z^2+1-2xz+2x-2z\)

\(A_{\left(x,z\right)}=\left(x^2-xz\right)+\left(z^2-xz\right)+\left(x-z\right)+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left[x\left(x-z\right)+z\left(z-x\right)+\left(x-z\right)\right]+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left[\left(x-z\right)\left(x-z+1\right)\right]+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left(x-z+1\right)\left(x-z+1\right)=\left(x-z+1\right)^2\)

Vậy nghiệm đã thức là: \(x-z+1=0\Leftrightarrow x-2y+1=0\)

p/s: lớp 8 không dài dòng thế này%