2) A = 3x² + y² - 2xy - 4x + 5. Tìm giá trị nhỏ nhất.( Helpppp mị đuyyyy!!!)

1) Ta có : 2n² - n - 1

= 2n² + 3n - 4n - 6 + 5

= ( 2n² + 3n ) - ( 4n + 6 ) + 5

= n( 2n + 3 ) - 2( 2n + 3 ) + 5

= ( 2n + 3 )( n - 2 ) + 5

Có ( 2n + 3 )( n - 2 ) chia hết cho ( 2n + 3 )

Để phép chia là phép chia hết thì 5 phải chia hết cho ( 2n + 3 )

hay ( 2n + 3 ) ∈ Ư(5) = { ±1 ; ±5 }

Vậy n ∈ { ±1 ; -2 ; -4 }

2) A = 3x² + y² - 2xy - 4x + 5

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

= ( x - y )² + 2( x² - 2x + 1 ) + 3

= ( x - y )² + 2( x - 1 )² + 3 ≥ 3 ∀ x, y

Dấu "=" xảy ra khi x = y = 1

=> MinA = 3 <=> x = y = 1

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne\frac{1}{3}\\x\ne3\end{cases}}\)

\(=\frac{2x^3-12x^2+5x^2+18x-30x+45}{3x^3-18x^2-x^2+27x+6x-9}\)

\(=\frac{\left(2x^3-12x^2+18x\right)+\left(5x^2-30x+45\right)}{\left(3x^3-18x^2+27x\right)-\left(x^2-6x+9\right)}\)

\(=\frac{2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)}{3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)}\)

\(=\frac{\left(x^2-6x+9\right)\left(2x+5\right)}{\left(x^2-6x+9\right)\left(3x-1\right)}\)

\(=\frac{2x+5}{3x-1}\)

\(A=27x^3-27x^2+18x-6=3\left(9x^3-9x^2+6x-2\right)\)

\(B=2x^3-x^2+5x+6=2x^3-2x^2+x^2-x+6x+6==\left(2x^2+x+6\right)\left(x-1\right)\)

\(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)

\(\Leftrightarrow2x^2+2y^2+z^2+25-6y-2xy-8x-2z\left(x-y\right)=0\)

\(\Leftrightarrow\left[x^2+y^2+z^2-2xy-2z\left(x-y\right)\right]+\left(x^2-8x+16\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow\left[\left(x^2-2xy+y^2\right)-2z\left(x-y\right)+z^2\right]+\left(x-4\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\left[\left(x-y\right)^2-2z\left(x-y\right)+z^2\right]+\left(x-4\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)

Vì \(\left(x-y-z\right)^2\ge0\), \(\left(x-4\right)^2\ge0\), \(\left(y-3\right)^2\ge0\)\(\forall x,y,z\)

\(\Rightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2\ge0\)\(\forall x,y,z\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-z=0\\x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=x-y\\x=4\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}z=4-3=1\\x=4\\y=3\end{cases}}\)

Vậy \(x=4\), \(y=3\), \(z=1\)