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

\(\Leftrightarrow M=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\left(\frac{1}{2}x^2-2x+2\right)+\left(\frac{1}{2}y^2-2y+2\right)-2\)

\(\Leftrightarrow M=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-2\right)^2+\frac{1}{2}\left(y-2\right)^2-2\ge-2\)\(\forall\)\(x\)

"=" khi x=y=2

Vậy Min M là -2 khi x=y=2

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

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

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

\(4M=\left[\left(2x-y\right)^2-2\left(2x-y\right)\times2+4\right]+3y^2-12y+4\)

\(4M=\left(2x-y-2\right)^2+3\left(y^2-4y+4\right)-8\)

\(4M=\left(2x-y-2\right)^2+3\left(y-2\right)^2-8\)

\(\Rightarrow4M\ge-8\)

\(\Leftrightarrow M\ge-2\)

Dấu "=" xảy ra khi :

\(m_1;c_1;\Delta t_1;t_1\):   nhôm           ;              \(m_2;c_2;\Delta t_2;t_2\):   nước

\(t_{cb}\): nhiệt độ cân bằng

\(m_1c_1\Delta t_1=m_2c_2\Delta t_2\)

\(\Leftrightarrow m_1c_1\left(t_1-t_{cb}\right)=m_2c_2\left(t_{cb}-t_2\right)\)

\(\Leftrightarrow m_1.880.\left(100-25\right)=47.4200.\left(25-20\right)\)

bn tự tính m1 nha

M=(x²-2x+1) + (y²-2y+1)-xy=(x-1)² + (y-1)²-xy

Đặt     \(x+\frac{1}{x}=t\)

\(\Rightarrow\)\(x^2+\frac{1}{x^2}=t^2-2\)

Khi đó pt trở thành:     \(t^2-2+3t+4=0\)

                         \(\Leftrightarrow\) \(t^2+3t+2=0\)

                         \(\Leftrightarrow\)   \(\left(t+1\right)\left(t+2\right)=0\)

                         \(\Leftrightarrow\)\(\orbr{\begin{cases}t+1=0\\t+2=0\end{cases}}\)

Thay trở lại ta có:     \(\orbr{\begin{cases}x+\frac{1}{x}+1=0\\x+\frac{1}{x}+2=0\end{cases}}\)

  TH1:    \(x+\frac{1}{x}+1=0\)           

      \(\Leftrightarrow\)\(\frac{x^2+1+x}{x}=0\)

      \(\Rightarrow\)  \(x^2+x+1=0\)

     \(\Leftrightarrow\) \(\left(x+0,5\right)^2+0,75=0\) 

      \(\Rightarrow\) pt vô nghiệm

TH2:   \(x+\frac{1}{x}+2=0\)

     \(\Leftrightarrow\) \(\frac{x^2+1+2x}{x}=0\)

     \(\Rightarrow\) \(x^2+2x+1=0\)

    \(\Leftrightarrow\) \(\left(x+1\right)^2=0\)

    \(\Leftrightarrow\) \(x+1=0\)

    \(\Leftrightarrow\) \(x=-1\)

Vậy...

Áp dụng BĐt bu-nhi-a , ta có \(\left(x+y\right)^2\le2\left(x^2+y^2\right)=6\)

Dấu = xảy ra <=> x=y=\(\sqrt{\frac{3}{2}}\) hoặc \(x=y=-\sqrt{\frac{3}{2}}\)

ban co cach khac ko

minh moi hoc lop 8 thoi