2.4:
a: \(A=x^2-4x+\frac92\)
\(=x^2-4x+4+\frac12\)
\(=\left(x-2\right)^2+\frac12\ge\frac12\forall x\)
Dấu '=' xảy ra khi x=2
b: \(B=2x^2+8x+10\)
\(=2\left(x^2+4x+5\right)\)
\(=2\left(x^2+4x+4+1\right)\)
\(=2\left(x+2\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x+2=0
=>x=-2
c: \(C=x^2+y^2-2xy+1\)
\(=x^2-2xy+y^2+1\)
\(=\left(x-y\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi x-y=0
=>x=y
d: \(D=x^2+y^2-2x-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}x-1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=2\end{cases}\)
Bài 2.3:
a: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
=>\(2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)
=>\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)
=>12x-5=0
=>12x=5
=>\(x=\frac{5}{12}\)
b: \(\left(x-5\right)^2-x\left(x-4\right)=9\)
=>\(x^2-10x+25-x^2+4x=9\)
=>-6x=9-25=-16
=>\(x=\frac{16}{6}=\frac83\)
c: \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)
=>\(x^2-10x+25-\left(x-4\right)\left(x-1\right)=0\)
=>\(x^2-10x+25-\left(x^2-5x+4\right)=0\)
=>\(x^2-10x+25-x^2+5x-4=0\)
=>-5x+21=0
=>-5x=-21
=>\(x=\frac{21}{5}\)
