hình như sai đề phải bn ???????????

Ko sai đâu bạn đề thi HSG Toán Tỉnh Lâm Đồng đó!

\(4A=12x^2+12y^2+4z^2+20xy-12yz-12zx-8x-8y+12\)

\(=9x^2+9y^2+4z^2+18xy-12yz-12zx+2\left(x^2+y^2+4-4x-4y+2xy\right)+x^2+y^2-2xy+4\)

\(=\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\)

Dấu \(=\)khi \(\hept{\begin{cases}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\).

Vậy \(minA=1\)khi \(x=y=1,z=3\).

\(A=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}\left(x^2y^2+\frac{2}{3}xy-\frac{8}{3}x-\frac{8}{3}y\right)+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}y^2-\frac{16}{9}y-\frac{16}{9}]\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2-\frac{2y}{9}]+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2]+1\)

\(\Leftrightarrow A\ge1\Leftrightarrow MinA=1\)

Dấu '' = '' xảy ra khi:

\(\hept{\begin{cases}z-\frac{3}{2}x-\frac{3}{2}y=0\\y-1=0\\x+\frac{y}{3}-\frac{4}{3}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}z=0\\y=1\\x=1\end{cases}}\)

22 nha

y=2 nha

A+4=mc2

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= 100000

đề bài này sai rồi ! 

Trả lời:

\(\left(-3xy^4+\frac{1}{2}x^2y^2\right)^3\)

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

\(=-27x^3y^{12}+3.9x^2y^8.\frac{1}{2}x^2y^2+3.\left(-3xy^4\right).\frac{1}{4}x^4y^4+\frac{1}{8}x^6y^6\)

\(=-27x^3y^{12}+\frac{27}{2}x^4y^{10}-\frac{9}{4}x^5y^8+\frac{1}{8}x^6y^6\)

\(x^4+6x^3+7x^2-6x+1=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)

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

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

a) \(x^2-xy+y^2=3\)

\(\Leftrightarrow\left(x+y\right)^2=3\left(1+xy\right)\)

\(\Rightarrow xy\ge-1\).

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

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

\(\Rightarrow xy\le-1\)

Do vai trò \(x,y\)như nhau nên giả sử \(x\ge y\).

- \(xy=-1\Rightarrow x=1,y=-1\).

Thử lại thỏa mãn. 

- \(xy=0\Rightarrow\orbr{\begin{cases}x=0\\y=0\end{cases}}\)dễ thấy đều không thỏa. 

- \(xy=1\Rightarrow\orbr{\begin{cases}x=y=1\\x=y=-1\end{cases}}\)không thỏa. 

- \(xy=2\Rightarrow\orbr{\begin{cases}x=2,y=1\\x=-1,y=-2\end{cases}}\)thỏa. 

- \(xy=3\Rightarrow\orbr{\begin{cases}x=3,y=1\\x=-1,y=-3\end{cases}}\)không thỏa. 

b) \(x=0\Rightarrow y=\pm1\)thỏa mãn. 

\(x\ne0\):

 \(y^2=1+x+x^2+x^3+x^4\)

\(\Leftrightarrow4y^2=4+4x+4x^2+4x^3+4x^4\)

Ta có: \(4x^4+4x^3+4x^2+4x+4>4x^4+4x^3+x^2=\left(2x^2+x\right)^2\)

\(4x^4+4x^3+4x^2+4x+4< 4x^4+4x^3+9x^2+4x+4=\left(2x^2+x+2\right)^2\)

suy ra \(4y^2=\left(2x^2+x+1\right)^2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Tử đây suy ra \(y\).

10 nha

kobit

hahaha