Question

Tim gia tri nho nhat cua bieu thuc A:

A=\(\frac{x^2+2x+3}{\left(x+2\right)^2}\)

Answer 1

ĐK : \(x\ne-2\)

ta có \(A=\frac{x^2+2x+3}{\left(x+2\right)^2}=\frac{3x^2+6x+9}{3\left(x+2\right)^2}=\frac{2x^2+8x+8+x^2-2x+1}{3\left(x+2\right)^2}\)

             \(=\frac{2\left(x+2\right)^2+\left(x-1\right)^2}{3\left(x+2\right)^2}=\frac{2}{3}+\frac{\left(x-1\right)^2}{3\left(x+2\right)^2}\) 

vì (x-1)^2 >=0=> \(\frac{\left(x-1\right)^2}{3\left(x+2\right)^2}>=0\)

=> \(A>=\frac{2}{3}\)

dấu = xảy ra <=> x=1 ( thỏa mãn ĐKXĐ)