\(A=\left(x+2\right)^2+\left|x+2\right|+15\)

Ta có:

\(\left(x+2\right)^2\ge0\forall x\)

\(\left|x+2\right|\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+\left|x+2\right|\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+\left|x+2\right|+15\ge15\forall x\)

\(\Rightarrow A\ge15\)Dấu bằng xảy ra.

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy \(minA=15\Leftrightarrow x=-2\)

\(A=x-x^2=-\left(x^2-x\right)=-\left(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}-\frac{1}{4}\right)=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Mình mới lớp 5 thôi

A=x(1-x)

gtln là 1

Vì | x -3 | > hoặc = 0

Suy ra : |x-3|+50 >hoặc =50

Vì A nhỏ nhất suy ra | x-3 | +50 =50

Suy ra x-3 =0

Suy ra x=3

Vậy GTNN của A = 50 khi x=3

a.

\(A=B\)

\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)

\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)

\(\Leftrightarrow8x+16=0\)

\(\Leftrightarrow8\left(x+2\right)=0\)

\(\Leftrightarrow x=-2\left(ktm\right)\)

Vậy không có giá trị x thỏa mãn A=B

b.

\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)

\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)

\(\Leftrightarrow\dfrac{8x}{-16}< 0\)

\(\Leftrightarrow\dfrac{8x}{16}>0\)

\(\Leftrightarrow\dfrac{x}{2}>0\)

\(\Leftrightarrow x>0\)