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a)(2x2+1)(3x3-2x2+3
= 6x5-4x4+6x2+3x3-2x2+3
= 6x5-4x4+3x3+4x2+3
b)(-3x+1)(4x4-x³+x)
= -12x5+3x4-3x2+4x4-x³+x
= -12x5+7x4-x3-3x2+x
a,\(8x^2-8xy+2x=2x\left(4x-8y+1\right)\)
b,\(\left(x^2+2x\right)\left(x^2+4x+3\right)-24=x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24=\left(t+1\right)\left(t-1\right)-24=t^2-5^2=\left(t+5\right)\left(t-5\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)( đặt t = x2 + 3x + 1 )
a) \(A=x-x^2=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy Max A = \(\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)
b) \(B=2x-2x^2=2\left(x-x^2\right)=-2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\le\frac{1}{2}\)
Vậy Max B = \(\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)
\(B=\left(x-3\right)^2+\left(x-11\right)^2\)
\(=x^2-6x+9+x^2-22x+121\)
\(=2\left(x^2-14x+49\right)+32\)
\(=2\left(x-7\right)^2+32\)
Ta có: \(\left(x-7\right)^2\ge0\Leftrightarrow2\left(x-7\right)^2+32\ge32\)
Vậy \(MinB=32\Leftrightarrow x=7\)
\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt \(c=x^2-5x\)lúc này \(C\)thành: \(C=\left(c-6\right)\left(c+6\right)=c^2-36\)
Mà: \(c^2\ge0\forall c\Leftrightarrow c^2-36\ge-36\Leftrightarrow C\ge-36\)
Dấu '' = '' xảy ra: \(c=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Vậy \(MinC=-36\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Bài làm
\(\left(x^2-2\right)\left(1-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)