\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

1)

\(A=-5x^2-4x+1\)

\(A=-5\left(x^2+\dfrac{4}{5}x-\dfrac{1}{5}\right)\)

\(A=-5\left(x^2+\dfrac{4}{5}x+\dfrac{4}{25}-\dfrac{9}{25}\right)\)

\(A=-5\left[\left(x+\dfrac{2}{5}\right)^2-\dfrac{9}{25}\right]\)

\(A=-\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{25}\le\dfrac{9}{25}\)

Dấu "=" xảy ra khi:

\(x=-\dfrac{2}{5}\)

2)

\(A=\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)\)

\(A=\left[\left(x-1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]\)

\(A=\left[x\left(x-8\right)-1\left(x-8\right)\right]\left[x\left(x-5\right)-4\left(x-5\right)\right]\)

\(A=\left(x^2-8x-x+8\right)\left(x^2-5x-4x+20\right)\)

\(A=\left(x^2-9x+8\right)\left(x^2-9x+20\right)\)

\(A=\left(x^2-9x+14-6\right)\left(x^2-9x+14+6\right)\)

\(A=\left(x^2-9x+14\right)^2-36\ge-36\)

Dấu "=" xảy ra khi:

\(x^2-9x+14=0\)

\(\Leftrightarrow x^2-2x-7x+14=0\)

\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)

Vậy...

\(A=−5x^2−4x+1 \)

=\(-5\left(x^2+\dfrac{4}{5}x-\dfrac{1}{5}\right)\)=\(-5\left(x^2+\dfrac{4}{5}+\dfrac{4}{25}-\dfrac{9}{25}\right)\)

=\(-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\)

Với mọi giá trị của x thì \(-5\left(x+\dfrac{2}{5}\right)^2\)nhỏ hơn hoặc bằng 0

=>\(\dfrac{9}{5}-5\left(x+\dfrac{2}{5}\right)^2\)nhỏ hơn hoặc bằng \(\dfrac{9}{5}\)

Hay Anhỏ hơn hoặc bằng \(\dfrac{9}{5}\)

Để A\(=\dfrac{9}{5}\)thì \(\left(x+\dfrac{2}{5}\right)^2=0\)

=>.\(x+\dfrac{2}{5}=0\)=>\(x=-\dfrac{2}{5}\)

Vậy ....

Theo mk câu 1 bác kia giải sai nhé

C=|2x-3/5|+4/3>=4/3

Dấu = xảy ra khi x=3/10

D=|x-3|+|-x-2|>=|x-3-x-2|=5

Dấu = xảy ra khi -2<=x<=3

\(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)

\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)

\(=\left(x^2+4x\right)^2-25\ge-25\)

\(\Rightarrow A_{min}=-25\)

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