Bài 5:
a: \(A=8x^2-8x+14\)
\(=8\left(x^2-x+\frac74\right)\)
\(=8\left(x^2-x+\frac14+\frac32\right)\)
\(=8\left(x-\frac12\right)^2+12\ge12\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
b: \(B=x^2+x+2\)
\(=x^2+x+\frac14+\frac74\)
\(=\left(x+\frac12\right)^2+\frac74\ge\frac74\forall x\)
Dấu '=' xảy ra khi \(x+\frac12=0\)
=>\(x=-\frac12\)
Bài 4:
a: \(A=-x^2-2x+5\)
\(=-\left(x^2+2x-5\right)\)
\(=-\left(x^2+2x+1-6\right)\)
\(=-\left(x+1\right)^2+6\le6\forall x\)
Dấu '=' xảy ra khi x+1=0
=>x=-1
b: \(B=9x-3x^2+4\)
\(=-3\left(x^2-3x-\frac43\right)\)
\(=-3\left(x^2-3x+\frac94-\frac94-\frac43\right)=-3\left(x-\frac32\right)^2+3\cdot\left(\frac94+\frac43\right)\)
\(=-3\left(x-\frac32\right)^2+3\cdot\left(\frac{27}{12}+\frac{16}{12}\right)=-3\left(x-\frac32\right)^2+\frac{43}{4}\le\frac{43}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\frac32=0\)
=>\(x=\frac32\)
Bài 3:
a: \(\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\)
=>\(x^2+4x+4+x^2+6x+9-2\left(x^2-5x+6\right)=19\)
=>\(2x^2+10x+13-2x^2+10x-12=19\)
=>20x+1=19
=>20x=18
=>\(x=\frac{18}{20}=\frac{9}{10}\)
b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\)
=>\(x^3+8-x^3+5x=15\)
=>5x=15-8=7
=>\(x=\frac75\)
