Câu hỏi của THN

Question

Tim gia tri nho nhat cua bieu thuc A:

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

vũ tiền châu · Accepted Answer

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)

Câu trả lời:

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)