\(A=x^2+10y^2+2x-6xy-10y+25\)

=> \(A=x^2+2x\left(1-3y\right)+\left(1-3y\right)^2-\left(1-3y\right)^2-10y+25\)

=> \(A=\left(x+1-3y\right)^2-1+6y-9y^2-10y+25\)

=> \(A=\left(x+1-3y\right)^2-9y^2-4y+24\)

=> \(A=\left(x+1-3y\right)^2-\left(3y\right)^2-2.3y.\frac{2}{3}-\left(\frac{2}{3}\right)^2+\frac{220}{9}\)

=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\)

Có \(\left(x+1-3y\right)^2\ge0\)với mọi x, y

\(\left(3y+\frac{2}{3}\right)^2\ge0\)với mọi y

=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\ge\frac{220}{9}\)với mọi x, y

Dấu "=" xảy ra <=> \(\left(x+1-3y\right)^2=0\)<=> \(x+1-3y=0\)

và \(\left(3y+\frac{2}{3}\right)^2=0\)=> \(3y+\frac{2}{3}=0\)

=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)

Bổ xung phần kết luận

KL: A_min = \(\frac{220}{9}\)<=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)

Mình đang cần gấp . Đảm bảo k trả đầy đủ + kb :'>

2.    \(Q=\left(x-3\right)\left(4x+5\right)+2019\)

        \(Q=4x^2+5x-12x-15+2019\)   

        \(Q=4x^2-7x+2004\)  

        \(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\) 

        \(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)  

        \(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)  

        \(\Rightarrow Q\ge\frac{32255}{16}\) 

         \(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)

3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)  

   \(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\) 

   \(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\)  (do a+b=1)

   \(T=4a^2-4ab+4a^2-6a^2-6b^2\) 

   \(T=-2a^2-4ab-2b^2\)

   \(T=-2\left(a^2+2ab+b^2\right)\) 

   \(T=-2\left(a+b\right)^2\)

   \(T=-2.1^2=-2.1=-2\) (do a+b=1)

\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

\(b,B=4x^2+4x-2\)

\(B=4x^2+4x+1-3\)

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

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

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

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

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

\(\Leftrightarrow x=-\frac{1}{2}\)