Ta có: \(A=\frac{x^2+x+1}{x^2-4x+4}\)

\(=\frac{x^2-4x+4+5x-3}{x^2-4x+4}=1+\frac{5x-10+7}{x^2-4x+4}\)

\(=1+\frac{5\left(x-2\right)}{\left(x-2\right)^2}+\frac{7}{\left(x-2\right)^2}=\frac{7}{\left(x-2\right)^2}+\frac{5}{x-2}+1\)

\(=7\left\lbrack\frac{1}{\left(x-2\right)^2}+\frac{1}{x-2}\cdot\frac57+\frac17\right\rbrack\)

\(=7\cdot\left\lbrack\frac{1}{\left(x-2\right)^2}+2\cdot\frac{1}{x-2}\cdot\frac{5}{14}+\frac{25}{196}+\frac{3}{196}\right\rbrack=7\left(\frac{1}{x-2}+\frac{5}{14}\right)^2+\frac{3}{28}\ge\frac{3}{28}\forall x\)

Dấu '=' xảy ra khi \(\frac{1}{x-2}+\frac{5}{14}=0\)

=>\(\frac{1}{x-2}=\frac{-5}{14}\)

=>x-2=-14/5

=>x=-4/5(nhận)

Đáp án đúng : A

\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)

\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{2}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)

\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)

bạn khùng đây là vật lý hỏ

\(A=7-x^2-3x=-\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{37}{4}=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)

\(maxA=\dfrac{37}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

\(A=-3\left(x^2-\dfrac{5}{3}x-2\right)=-3\left(x^2-2\cdot\dfrac{5}{6}x+\dfrac{25}{36}-\dfrac{97}{36}\right)\\ A=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\\ A_{max}=\dfrac{97}{12}\Leftrightarrow x=\dfrac{5}{6}\)

A= 3x² - 2x + 3

= 3(x²- 2/3x + 1/9 ) + 8/3

= 3(x-1/3)² + 8/3 > 8/3 \(\forall\)x

dấu ''='' xảy ra <=> x = 1/3

/HT\

Nhầm đề rồi mấy bạn trả lời

Bảo là giá trị nguyên của ,\(\frac{2x-3}{3x+2}\) , các bạn ghi là \(3x^2-2x+3\)rồi

HT