\(A = - x^{2} - 7 y^{2} - 4 x y + 16 y + 2 x - 12\) \(= - \left(\right. x^{2} + 4 x y + 7 y^{2} \left.\right) + 2 x + 16 y - 12\) \(= - \left(\right. \left(\right. x + 2 y \left.\right)^{2} + 3 y^{2} \left.\right) + 2 x + 16 y - 12\)

ta có

\(u = x + 2 y \Rightarrow x = u - 2 y\)

ta thay

\(A = - \left(\right. u - 1 \left.\right)^{2} - 3 \left(\right. y - 2 \left.\right)^{2} + 1\)

Vì \(\left(\right. u - 1 \left.\right)^{2} \geq 0 , \left(\right. y - 2 \left.\right)^{2} \geq 0 \Rightarrow A \leq 1\)

ta có

\(\textrm{ }u=1,\textrm{ }y=2\textrm{ }\Rightarrow x=-3\)

vậy

\(max⁡A=1\text{t}ạ\text{i}\left(\right.x,y\left.\right)=\left(\right.-3,2\left.\right)\)

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

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$\frac{47}{15}(3x^2+5y^2)=[(\sqrt{3}x)^2+(-\sqrt{5}y)^2][(\frac{2}{\sqrt{3}})^2+(\frac{3}{\sqrt{5}})^2]\geq (2x-3y)^2$

$\Leftrightarrow \frac{47}{15}(3x^2+5y^2)\geq 49$

$\Rightarrow 3x^2+5y^2\geq \frac{735}{47}$

Ta có đpcm.

a)

Ta có:

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

\(\ge0-2=-2\)

Vậy \(A_{min}=-2\), đạt được khi và chỉ khi \(x-1=0\Leftrightarrow x=1\)

b)\(B=4x^2+4x+8=4x^2+4x+1+7\)

\(=\left(2x+1\right)^2+7\ge0+7=7\)

Vậy \(B_{min}=7\), đạt được khi và chỉ khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)

c)

Ta có:

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

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

\(=\dfrac{17}{4}-\left(x-\dfrac{3}{2}\right)^2\le\dfrac{17}{4}-0=\dfrac{17}{4}\)

Vậy \(C_{max}=\dfrac{17}{4}\), đạt được khi và chỉ khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

d) Ta có:

\(D=-x^2-5x=-\left(x^2+5x\right)=\dfrac{25}{4}-\left(x^2+2x.\dfrac{5}{2}+\dfrac{25}{4}\right)\)

\(=\dfrac{25}{4}-\left(x+\dfrac{5}{2}\right)^2\le\dfrac{25}{4}-0=\dfrac{25}{4}\)

Vậy \(D_{max}=\dfrac{25}{4}\), đạt được khi và chỉ khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)

e) Ta có:

\(E=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2+4y^2+5^2-4xy+10x-20y+y^2-2y+1+2\)

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

\(\ge0+0+2=2\)

Vậy \(E_{min}=2\), đạt được khi và chỉ khi \(x-2y+5=y-1=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

= 2\(x^{2}\) + 3y\(^{2}\) - 8\(x\) - 6y + 15

A = 2(\(x^{2}\) - 4\(x\) + 4) + 3(y\(^{2} - 2 y + 1\)) + 6

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

Vì (\(x - 2 \left.\right)^{2}\) ≥ 0; ∀ \(x\); (y -1)\(^{2}\) ≥ 0 ∀ y 

⇒ 2.(\(x - 2 \left.\right)^{2}\) ≥ 0 ∀ \(x\); 3(y - 1)\(^{2}\) + 4 ≥ y ∀ y 

2.(\(x - 2 \left.\right)^{2}\) + 3(y - 1)\(^{2}\) + 4 ≥ 4; Dấu bằng xảy ra khi:

\(\left{\right. x - 2 = 0 \\ y - 1 = 0\)

\(\left{\right. x = 2 \\ y = 1\)

Vậy A đạt giá trị nhỏ nhất là 4 khi (\(x ; y\)) = (2; 1) 

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

2:

a: \(3xy^2-3x^3-6xy+3x\)

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

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

=3x(y-1-x)(y-1+x)

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

\(=x\left(16+4xy-4y^2-x^2\right)\)

\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

=x(4-x+2y)(4+x-2y)

d: \(xz-x^2-yz+2xy-y^2\)

=z(x-y)-\(\left(x^2-2xy+y^2\right)\)

=\(z\left(x-y\right)-\left(x-y\right)^2\)

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

=(2x-y)(2x+y)-3(2x-y)

=(2x-y)(2x+y-3)

f: \(x^4-x^3-10x^2+2x+4\)

\(=x^4+2x^3-2x^2-3x^3-6x^2+6x-2x^2-4x+4\)

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

g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)

\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)

\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)

=(x-1)(x^2+1)(11-5y-1)(11+5y+1)

=(x-1)(x^2+1)(10-5y)(12+5y)

=5(2-y)(x-1)(x^2+1)(5y+12)