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

\(A=x^2-4xy+4y^2+x^2+2x+1+2018\)

\(A=\left(x-2y\right)^2+\left(x+1\right)^2+2018\ge2018\)

\(A_{min}=2018\) khi \(\left\{{}\begin{matrix}x=-1\\y=-\frac{1}{2}\end{matrix}\right.\)

\(B=-\left(4x^2+4xy+y^2\right)-\left(x^2-6x+9\right)+2029\)

\(B=-\left(2x+y\right)^2-\left(x-3\right)^2+2029\le2029\)

\(B_{max}=2029\) khi \(\left\{{}\begin{matrix}x=3\\y=-6\end{matrix}\right.\)

a)

$x^2-2x+5y^2-4y+2020=(x^2-2x+1)+5(y^2-\frac{4}{5}y+\frac{2^2}{5^2})+\frac{10091}{5}$

$=(x-1)^2+5(y-\frac{2}{5})^2+\frac{10091}{5}$

$\geq \frac{10091}{5}$

Vậy GTNN của biểu thức là $\frac{10091}{5}$. Giá trị này đạt được tại $(x-1)^2=(y-\frac{2}{5})^2=0$

$\Leftrightarrow x=1; y=\frac{2}{5}$

b)

\(B=(x-5)^2-(3x-7)^2=(x-5-3x+7)(x-5+3x-7)\)

\(=(2-2x)(4x-12)=8(1-x)(x-3)=8(x-3-x^2+3x)\)

\(=8(4x-3-x^2)=8[1-(x^2-4x+4)]=8[1-(x-2)^2]\)

Vì $(x-2)^2\geq 0, \forall x\in\mathbb{R}$ nên $1-(x-2)^2\leq 1$

$\Rightarrow B=8[1-(x-2)^2]\leq 8$. Vậy GTLN của biểu thức là $8$ khi $x=2$

c)

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

$=7-(x^2-2x+1)-(9y^2+6y+1)=7-(x-1)^2-(3y+1)^2$

Vì $(x-1)^2\geq 0; (3y+1)^2\geq 0$ với mọi $x,y$ nên $C=7-(x-1)^2-(3y+1)^2\leq 7$

Vậy GTLN của $C$ là $7$. Giá trị này đạt được tại $(x-1)^2=(3y+1)^2=0$

$\Leftrightarrow x=1; y=\frac{-1}{3}$

d)

$D=-5x^2-9y^2-7x+18y-2015=-(5x^2+7x)-(9y^2-18y)-2015$

$=-5(x^2+\frac{7}{5}x+\frac{7^2}{10^2})-9(y^2-2y+1)-\frac{40071}{20}$
$=-5(x+\frac{7}{10})^2-9(y-1)^2-\frac{40071}{20}$

$\leq -\frac{40071}{20}$

Vậy GTLN của biểu thức là $\frac{-40071}{20}$ khi $x=-\frac{-7}{10}; y=1$

e)

$E=5x^2+y^2-4xy+18x-4y+28=x^2+(4x^2+y^2-4xy)+18x-4y+28$

$=x^2+(2x-y)^2+4(2x-y)+10x+28$

$=(x^2+10x+25)+(2x-y)^2+4(2x-y)+4-1$

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

Vậy GTNN của $E$ là $-1$. Giá trị này xác định tại \(\left\{\begin{matrix} (x+5)^2=0\\ (2x-y+2)^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-5\\ y=-8\end{matrix}\right.\)

f)

$F=x^4+x^2-6x+9=(x^4-2x^2+1)+(3x^2-6x+3)+5$
$=(x^4-2x^2+1)+3(x^2-2x+1)+5$

$=(x^2-1)^2+3(x-1)^2+5$

$\geq 5$

Vậy GTNN của $F$ là $5$. Giá trị này đạt được khi $(x^2-1)^2=(x-1)^2=0$

$\Leftrightarrow x=1$

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

d)  D = x⁴ - 6x² + 10

D = (X²)² - 2. x². 3 + 3² + 1

D = (x² - 3)² + 1

(x² - 3)²  >= 0 với mọi x

(x² - 3)² + 1 >=1 với moi5 x

Vậy GTNN của D là 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 \(x=-\dfrac{5}{4}\)

Lời giải:

ĐKĐB $\Leftrightarrow (x^2+4y^2-4xy)+8x=5$

$\Leftrightarrow (x-2y)^2+8x=5$.

Đặt $x-2y=a; x=b$ thì bài toán trở thành:

Cho $a,b$ thực thỏa mãn $a^2+8b=5$. Tìm max của $B=-2a+8b$

Áp dụng BĐT AM-GM:

$a^2+1\geq 2\sqrt{a^2}=2|a|\geq -2a$

$\Rightarrow a^2+1\geq -2a$

$\Rightarrow a^2+8b+1\geq -2a+8b$

$\Leftrightarrow 6\geq B$. Vậy $B_{\max}=6$