g: a+b=2
=>a=2-b
\(G=a^2+b^2\)
\(=\left(2-b\right)^2+b^2\)
\(=b^2-4b+4+b^2\)
\(=2b^2-4b+4=2\left(b^2-2b+2\right)\)
\(=2\left(b^2-2b+1+1\right)=2\left(b-1\right)^2+2\ge2\forall b\)
Dấu '=' xảy ra khi b-1=0
=>b=1
=>a=2-b=2-1=1
h: \(H=\left(x-2\right)\left(x-3\right)\left(x-6\right)\left(x+1\right)\)
\(=\left(x^2-3x-2x+6\right)\left(x^2+x-6x-6\right)\)
\(=\left(x^2-5x+6\right)\left(x^2-5x-6\right)\)
\(=\left(x^2-5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi \(x^2-5x=0\)
=>x(x-5)=0
=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)
i: \(I=\left(x^2+4x+5\right)\left(x^2+4x+6\right)+3\)
\(=\left(x^2+4x+5\right)\left(x^2+4x+5+1\right)+3\)
\(=\left(x^2+4x+5\right)^2+\left(x^2+4x+5\right)+3\)
\(=\left(x^2+4x+5\right)^2+2\cdot\left(x^2+4x+5\right)\cdot\frac12+\frac14+\frac{11}{4}\)
\(=\left(x^2+4x+5+\frac12\right)^2+\frac{11}{4}\)
\(=\left(x^2+4x+4+\frac32\right)^2+\frac{11}{4}\)
\(=\left\lbrack\left(x+2\right)^2+\frac32\right\rbrack^2+\frac{11}{4}\ge\left(\frac32\right)^2+\frac{11}{4}=\frac94+\frac{11}{4}=\frac{20}{4}=5\forall x\)
Dấu '=' xảy ra khi x+2=0
=>x=-2
k: \(K=x^2+2y^2+2xy+2x+4y+2025\)
\(=x^2+2xy+y^2+2x+2y+y^2+2y+1+2024\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y+1\right)^2+2023\)
\(=\left(x+y+1\right)^2+\left(y+1\right)^2+2023\ge2023\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+1=0\\ x+y+1=0\end{cases}\Rightarrow\begin{cases}y=-1\\ x=-\left(y+1\right)=-\left(-1+1\right)=0\end{cases}\)
