Bài 1:
\(a,A=4x^2-12x+10\\ =\left(4x^2-12x+9\right)+1\\ =\left(2x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra: `2x-3=0<=>x=3/2`
\(b,B=2x-x^2-2\\ =\left(-x^2+2x-1\right)-1\\ =-\left(x^2-2x+1\right)-1\\ =-\left(x-1\right)^2-1\le-1\forall x\)
Dấu "=" xảy ra: `x-1=0<=>x=1`
\(c,C=4x-x^2+3\\ =\left(-x^2+4x-4\right)+7\\ =-\left(x^2-4x+4\right)+7\\ =-\left(x-2\right)^2+7\le7\forall x\)
Dấu "=" xảy ra: `x-2=0<=>x=2`
\(d,D=x-x^2\\ =\left(-x^2+x-\dfrac{1}{4}\right)+\dfrac{1}{4}\\ =-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra: `x-1/2=0<=>x=1/2`
\(e,E=2x^2-8x-10\\ =\left(2x^2-8x+8\right)-18\\ =2\left(x^2-4x+4\right)-18\\ =2\left(x-2\right)^2-18\ge-18\forall x\)
Dấu "=" xảy ra: `x-2=0<=>x=2`
Bài 2:
a: \(x^2-y^2=\left(x-y\right)\left(x+y\right)=\left(87-13\right)\left(87+13\right)\)
\(=74\cdot100=7400\)
b: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
\(=\left(101-1\right)^3=100^3=1000000\)
c: \(x^3+9x^2+27x+27\)
\(=x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3\)
\(=\left(x+3\right)^3=\left(97+3\right)^3=100^3=1000000\)
