a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng

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

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

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

MinM=-4

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x-y=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

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

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

\(=4x^2-7x+2\)

\(=\left(2x\right)^2-2\cdot2\cdot\frac{7}{4}x+\left(\frac{7}{4}\right)^2-\frac{17}{16}\)

\(=\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\)

Vì \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\ge-\frac{17}{16}\forall x\)

Dấu " = " xảy ra khi và chỉ khi \(\left(2x-\frac{7}{4}\right)^2=0\)

\(\Leftrightarrow x=\frac{7}{8}\)

Vậy \(H_{min}=-\frac{17}{16}\)tại \(x=\frac{7}{8}\)

\(x^2+\left(x-2\right)\left(3x-1\right)=x^2+3x^2-x-6x+2=4x^2-7x+2\)

\(=4x^2-7x+\frac{49}{16}-\frac{17}{16}\)

\(=\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\)

Vì: \(\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\ge\frac{17}{16}\forall x\)

=> Min H =17/16 tại \(\left(2x-\frac{7}{4}\right)^2=0\Rightarrow x=\frac{7}{8}\)

=.= hok tốt!!

\(H=x^2+\left(x-2\right)\left(3x-1\right)=x^2+3x^2-x-6x+2=4x^2-7x+2\)

                               \(=4x^2-2.2x.\frac{7}{4}+\frac{49}{16}-\frac{17}{16}=\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\ge\frac{-17}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow2x-\frac{7}{4}=0\Leftrightarrow x=\frac{7}{8}\)

Vậy H_Min = -17/16 khi và chỉ khi x = 7/8

'=Bài 3:

\(Y=\left(x^{100}+1+1+1+1+1+1+1+1+1\right)-10x^{10}+1\)

Áp dụng BĐT Cauchy cho 10 số không âm ta có:

\(x^{100}+1+1+1+1+1+1+1+1+1\ge10\sqrt{x^{100}}=10x^{10}\)

\(Y\ge10x^{10}-10x^{10}+1=1\)

\(\Rightarrow maxY=1\)

Dấu "=" xảy ra\(\Leftrightarrow x^{100}=1\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

\(5x^2-4xy+y^2-4x+4=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(2x-y\right)^2+\left(x-2\right)^2=0\)

Do \(\left(2x-y\right)^2,\left(x-2\right)^2\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

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

\(=1+6^3=217\)

MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA

Ta có: \(4\ge2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

    \(\Rightarrow x+y\le2\)

Ta có: \(P=\sqrt{x\left(14x+10y\right)}+\sqrt{y\left(14y+10x\right)}\)

              \(=\sqrt{\dfrac{24x\left(14x+10y\right)}{24}}+\sqrt{\dfrac{24y\left(14y+10x\right)}{24}}\le\dfrac{\dfrac{24x+14x+10y}{2}}{\sqrt{24}}+\dfrac{\dfrac{24y+14y+10x}{2}}{\sqrt{24}}\)

\(\Leftrightarrow P\le\dfrac{24\left(x+y\right)}{2\sqrt{6}}\le\dfrac{24.2}{2\sqrt{6}}=4\sqrt{6}\)

Dấu "=" xảy ra ⇔ x = y = 1