\(ĐKXĐ:x\ne\frac{1}{2}\)

\(pt\Leftrightarrow2x^2+x>\left(1-2x\right)\left(1-x\right)\)

\(\Leftrightarrow2x^2+x>2x^2-3x+1\)

\(\Leftrightarrow x>-3x+1\)

\(\Leftrightarrow4x>1\Leftrightarrow x>\frac{1}{4}\)

Ta có:

A = 2x² + 2xy + y² - 2x + 2y + 2

A = (x² + 2xy + y²) + 2(x + y) + 1 + (x² - 4x + 4) - 3

A = (x + y)² + 2(x + y) + 1 + (x - 2)² - 3

A = (x + y + 1)² + (x - 2)² - 3 \(\ge\)-3 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x-1\\x=2\end{cases}}\) <=> \(\hept{\begin{cases}y=-2-1=-3\\x=2\end{cases}}\)

Vậy MinA = -3 <=> x = 2 và y = -3

\(2x^2+2xy+y^2-2x+2y+\)\(2\)

\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4x+2\right)-1\)

\(=\left(x+y+1\right)^2+\left(x-2\right)^2-1\)

Ta thấy \(\left(x+y+1\right)^2\ge0\)     \(\forall x,y\)

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

=> \(\left(x+y+1\right)^2+\left(x-2\right)^2\ge0\)     \(\forall x,y\)

=> \(\left(x+y+1\right)^2+\left(x-2\right)^2-1\ge-1\)

hay \(A\ge-1\)

\(MinA=-1\)\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Xét x không âm thì pt\(\Leftrightarrow3x^2-7x+4=0\)

\(\Leftrightarrow3x^2-3x-4x+4=0\)

\(\Leftrightarrow3x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)

Xét x âm thì pt\(\Leftrightarrow3x^2+7x+4=0\)

\(\Leftrightarrow3x^2+3x+4x+4=0\)

\(\Leftrightarrow3x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-4}{3}\\x=-1\end{cases}}\)

x=3-7+4=0

x=0

\(ĐKXĐ:x\ne2\)

\(pt\Leftrightarrow2x-5>x-2\)

\(\Leftrightarrow2x-x>5-2\)

\(\Leftrightarrow x>3\)

Vậy x > 3 

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

=> \(2x-5>-\left(2-x\right)\)

=> \(2x-5>x-2\)

=> \(2x-x>5-2\)

=> \(x>3\)

\(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3\ge a^2b+ab^2\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) ( luôn đúng )

BĐT \(\Leftrightarrow\left(\frac{a}{b}\right)^3+1\ge\frac{a}{b}\left(\frac{a}{b}+1\right)\) (chia hai vế cho b³ > 0)

Đặt \(\frac{a}{b}=t>0\). Ta cần chứng minh \(t^3+1\ge t\left(t+1\right)\Leftrightarrow\left(t-1\right)^2\left(t+1\right)\ge0\)

P/s: Làm kiểu khác cho nó lạ xíu:D