\(A=2x^2+6x\)
\(A=2\left(x^2+3x\right)\)
\(A=2\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(A=2\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}.2\)
\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{9}{2}\)
Dấu "=" xảy ra khi :
\(x=-\dfrac{3}{2}\)
B đã sửa đề vì theo đề của you thì ko có tổng nào = nhau
\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(B=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(B=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]\)
\(B=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)
\(B=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
\(B=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)\)
\(B=\left(x^2+5x+5\right)^2-1\ge-1\)
Dấu "=" xảy ra khi:
\(x^2+5x+5=0\)
Cạn-
