Tham khảo:

Câu hỏi của Nguyễn Tấn Phát

link:https://olm.vn/hoi-dap/detail/214667445437.html

ib đưa link

Mình cảm ơn ạ

\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu \("="\Leftrightarrow x=y=z=1\)

Em cảm ơn anh ạ! 

Anh giúp em ạ! 

https://hoc24.vn/cau-hoi/cho-abc-la-cac-so-duong-cmr-dfraca2bcdfracb2cadfracc2abgedfracabc2.4139278814936

Ta có:

\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2+2zx+x^2\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)+z^2=0\)\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2+\left(x+5\right)^2+\left(y+3\right)^2+z^2=0\)

Không tồn tại x,y,z thỏa mãn đề bài

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

\(\Rightarrow P\ge\dfrac{x+2y}{3x+y+5z}+\dfrac{y+2z}{3y+z+5x}+\dfrac{z+2x}{3x+x+5y}\)

\(\Rightarrow P\ge\dfrac{\left(x+2y\right)^2}{\left(x+2y\right)\left(3x+y+5z\right)}+\dfrac{\left(y+2z\right)^2}{\left(y+2z\right)\left(3y+z+5x\right)}+\dfrac{\left(z+2x\right)^2}{\left(z+2x\right)\left(3x+x+5y\right)}\)

\(\Rightarrow P\ge\dfrac{\left(x+2y\right)^2}{3x^2+2y^2+7xy+5xz+10yz}+\dfrac{\left(y+2z\right)^2}{3y^2+2z^2+7yz+5xy+10xz}+\dfrac{\left(z+2x\right)^2}{3z^2+2x^2+7xz+5yz+10xy}\)

\(\Rightarrow P\ge\dfrac{\left(x+2y+y+2z+z+2x\right)^2}{5\left(x^2+y^2+z^2\right)+22\left(xy+xz+yz\right)}\)

\(\Rightarrow P\ge\dfrac{9\left(x+y+z\right)^2}{5\left(x+y+z\right)^2+12\left(xy+xz+yz\right)}\ge\dfrac{9\left(x+y+z\right)^2}{5\left(x+y+z\right)^2+\dfrac{12\left(x+y+z\right)^2}{3}}\)

\(\Rightarrow P\ge1\)

\(\Rightarrow P_{min}=1\) khi \(x=y=z\)

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

b: \(B=-x^2-y^2+xy+2x+2y\)

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

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

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)

\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

b: \(B=-x^2-y^2+xy+2x+2y\)

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

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

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)

\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn