Ta có : A = x(x + 1)(x2 + x - 4)
= (x2 + x)(x2 + x - 4)
Đặt x2 + x = t
Khi đó A = t(t - 4)
= t2 - 4t = t2 - 4t + 4 - 4 = (t - 2)2 - 4 \(\ge\)-4
Dấu "=" xảy ra <=> t - 2 = 0
=> t = 2
=> x2 + x = 2
=> x2 + x - 2 = 0
=> x2 + 2x - x - 2 = 0
=> x(x + 2) - (x + 2) = 0
=> (x - 1)(x + 2) = 0
=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
Vậy Min A = -4 <=> x \(\in\left\{1;-2\right\}\)
A = x( x + 1 )( x2 + x - 4 )
= ( x2 + x )( x2 + x - 4 )
Đặt t = x2 + x
A <=> t( t - 4 )
= t2 - 4t
= ( t2 - 4t + 4 ) - 4
= ( t - 2 )2 - 4
= ( x2 + x - 2 )2 - 4 ≥ -4 ∀ x
Đẳng thức xảy ra <=> x2 + x - 2 = 0
<=> x2 - x + 2x - 2 = 0
<=> x( x - 1 ) + 2( x - 1 ) = 0
<=> ( x - 1 )( x + 2 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
=> MinA = -4 <=> x = 1 hoặc x = -2
a,\(A=x\left(x+1\right)\left(x^2+x-4\right)\)
\(=\left(x^2+x\right)\left(x^2+x-4\right)\)
Đặt \(x^2+x=t\)ta có:
\(A=t\left(t-4\right)\)
\(=t^2-4t\)
\(=\left(t^2-4t+4\right)-4\)
\(=\left(t-2\right)^2-4\ge-4\forall t\)
Dấu "="xảy ra khi \(\left(t-2\right)^2=0\Rightarrow t=2\)
\(\Rightarrow Min_A=-4\Leftrightarrow t=2\)
\(\Leftrightarrow x^2+x=2\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=1;x=2\)
b,\(B=-x^2-y^2+xy+2x+2y\)
\(\Leftrightarrow-2B=2x^2+2y^2-2xy-4x-4y\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)
\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge-8\Leftrightarrow B\le4\)
Dấu"="xảy ra khi \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-2\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow x=y=2}\)
Vậy \(Max_B=4\Leftrightarrow x=y=2\)
a) \(A=x.\left(x+1\right)\left(x^2+x-4\right)\)
\(=\left(x^2+x\right)\left(x^2+x-4\right)\)
\(=\left(x^2+x\right)^2-4.\left(x^2+x\right)\)
\(=\left(x^2+x\right)^2-2.\left(x^2+x\right).2+4-4\)
\(=\left(x^2+x-2\right)^2-4\ge-4\)
Hay \(A\ge-\frac{16}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x^2+x-2=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
Vậy GTNN của \(A=-4\) khi \(x=-2\) hoặc \(x=1\)
b) \(B=-x^2-y^2+xy+2x+2y\)
\(\Rightarrow-4B=4x^2+4y^2+4xy+8y+8x\)
\(=\left(x^2+4xy+4y^2\right)+3x^2+8.\left(x+y\right)\)
\(=\left(x+2y\right)^2+4.\left(x+2y\right)+3x^2+4x\)
\(=\left(x+2y\right)^2+2.\left(x+2y\right).2+4+3x^2+4x-4\)
\(=\left(x+2y+2\right)^2+3.\left(x^2+\frac{4}{3}x\right)-4\)
\(=\left(x+2y+2\right)^2+3.\left(x+\frac{2}{3}\right)^2-\frac{16}{3}\ge-\frac{16}{3}\)
Do đó : \(-4B\ge-\frac{16}{3}\Rightarrow B\le\frac{4}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2y+2=0\\x+\frac{2}{3}=0\end{cases}\Leftrightarrow}x=y=-\frac{2}{3}\)
Vậy GTLN của \(B=\frac{4}{3}\) khi \(x=y=-\frac{2}{3}\)
