Ta có \(3-\frac{8}{x^3}\)

\(=0=3-\frac{8}{x^3}=0=x=\frac{2}{^3\sqrt{3}}=y=\frac{9}{^3\sqrt{3}}=3^3\sqrt{9}\)

Vậy \(min\)của hàm số \(3x^2+\frac{4}{x}=3^3\sqrt{9}\)

Áp dụng hằng đẳng thức: \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)\) ta có: 

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow\left(x+1-x+1\right)^3+3\left(x+1\right)\left(x-1\right)\left(x+1-x+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2^3+6\left(x^2-1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow6\left[x^2-1-\left(x-1\right)^2\right]=-10-8\)

\(\Leftrightarrow6\left[\left(x-x+1\right)\left(x+x-1\right)-1\right]=-18\)

\(\Leftrightarrow2x-1-1=-3\)\(\Leftrightarrow2x-2=-3\)\(\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(x=\frac{-1}{2}\)

hahahahaha

\(ĐKXĐ:x\ne2\)

\(\frac{2x-5}{5x-10}=0\)\(\Leftrightarrow2x-5=0\)\(\Leftrightarrow2x=5\)\(\Leftrightarrow x=\frac{5}{2}\left(tmđk\right)\)

Vậy \(x=\frac{5}{2}\)

Ta có: A = 4 - 5x² - y² + 2xy - 4x

A = -(5x² + y² .- 2xy + 4x - 4)

A = -[(x² - 2xy + y²) + (4x² + 4x + 1) - 5]

A = -(x - y)² - (2x + 1)² + 5 \(\ge\)5 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\2x+1=0\end{cases}}\) => \(x=y=-\frac{1}{2}\)

Vậy MaxA = 5 <=> x = y = -1/2

3x-2x-2/x-2

=x-1/x-1

= 1

\( {{3x} \over x-1}\)+\({{-2x-1} \over x-1}\)=\({{3x+(-2x)-1} \over x-1}\)=\({{x-1} \over x-1}\)=1