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

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

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

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

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

\(2017-4x-x^2=2021-\left(x^2+4x+4\right)=2021-\left(x+2\right)^2\le2021\)

dấu "=" xảy ra khi x=-2

vậy gtln của biểu thức là 2021 khi x=-2

\(A=-\left(4x^2-4x+1\right)-\left(y^2+6y+9\right)+11\\ A=-\left(2x-1\right)^2-\left(y+3\right)^2+11\le11\\ A_{max}=11\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

a) \(C=4x^2+3y^2+4xy-4x-10y+7=\left[4x^2+4x\left(y-1\right)+\left(y-1\right)^2\right]+2\left(y^2-4y+4\right)-2=\left(2x+y-1\right)^2+2\left(y-2\right)^2-2\ge-2\)

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

d) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

Ta có: l x+1l lớn hơn hoặc bằng 0, với mọi x

          l 2y -3l lớn hơn hoặc bằng 0, với mọi y

=> l x+1l + l 2y-3l lớn hơn hoặc bằng 0, với mọi x,y

=> l x+1l + l 2y-3l + 5 lớn hơn hoặc bằng 5

=> 1/ lx+1l + l2y-3l + 5 bé hơn hoặc bằng 1/5

=> 20/ lx+1l + l2y-3l+5 bé hơn hoặc bằng 20/5 = 4

Vậy max Q = 4

Dẫu "=" xảy ra <=> x = -1 ; y = 3/2

Chúc bạn học tốt!

Ta có: l x+1l lớn hơn hoặc bằng 0, với mọi x

          l 2y -3l lớn hơn hoặc bằng 0, với mọi y

=> l x+1l + l 2y-3l lớn hơn hoặc bằng 0, với mọi x,y

=> l x+1l + l 2y-3l + 5 lớn hơn hoặc bằng 5

=> 1/ lx+1l + l2y-3l + 5 bé hơn hoặc bằng 1/5

=> 20/ lx+1l + l2y-3l+5 bé hơn hoặc bằng 20/5 = 4

Vậy max Q = 4

Dẫu "=" xảy ra <=> x = -1 ; y = 3/2

\(A=\dfrac{\sqrt{x-9}}{5x}\left(ĐKx\ge9\right)\)

A'=\(\dfrac{\dfrac{5x}{2\sqrt{x-9}}-5\sqrt{x-9}}{\left(5x^2\right)}\)

\(A'=0\rightarrow5x=10\left(x-9\right)\)

            \(\rightarrow x=18\)

\(MaxA=\dfrac{1}{30}\) khi \(x=18\)

\(A=\dfrac{2.3\sqrt{x-9}}{30x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)

\(A_{max}=\dfrac{1}{30}\) khi \(\sqrt{x-9}=3\Leftrightarrow x=18\)

E = - ( x² + 4x + 4 + y² - 2y + 1 - 5)

= -(x+2)² - (y-1)² + 5 \(\ge\)5

vậy Min đề = 5 khi -(x+2)² = 0 => -x - 2 = 0 => -x = 2 => x = -2 

và -(y-1)² = 0 => -y + 1 = 0 => -y = -1 => y = 1 

Iruko t lm nhầm 1 chút thui mak lm j zữ z , khinh ng =.=