1.(\(x^2-x+3\))(\(x^2-x-2\))\(+4\)
2.(\(x^2+5x+6\)) (\(x^2-15x+56\))\(-144\)
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1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x=t\)
\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)
\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)
Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)
\(\Delta=7^2-4.11=5\)
\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)
2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x=t\)
\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)
\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)
Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)
\(\Delta=\left(-8\right)^2-4.11=20\)
\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)
a)
\((6x+5)^2(3x+2)(x+1)-35\)
\(=(36x^2+60x+25)(3x^2+5x+2)-35\)
\(=[12(3x^2+5x+2)+1](3x^2+5x+2)-35\)
\(=(12a+1)a-35=12a^2+a-35\) (đặt \(3x^2+5x+2=a)\)
\(=4a(3a-5)+7(3a-5)=(4a+7)(3a-5)\)
\(=(12x^2+20x+15)(9x^2+15x+1)\)
b)
\(8(4x+1)(2x-3)(4x-3)(x+1)-130\)
\(=8[(4x+1)(4x-3)][(2x-3)(x+1)]-130\)
\(=8(16x^2-8x-3)(2x^2-x-3)-130\)
\(=8(8a+21)a-130\) (Đặt \(2x^2-x-3=a\) )
\(=64a^2+168a-130=2(8a-5)(4a+13)\)
\(=2(8x^2-4x+1)(16x^2-8x-29)\)
c)
\((4x+1)(12x-1)(3x+2)(x+1)-4\)
\(=[(4x+1)(3x+2)][(12x-1)(x+1)]-4\)
\(=(12x^2+11x+2)(12x^2+11x-1)-4\)
\(=(a+2)(a-1)-4\) (đặt \(a=12x^2+11x\) )
\(=a^2+a-6=(a-2)(a+3)\)
\(=(12x^2+11x-2)(12x^2+11x+3)\)
d)
\((x+2)(x+3)^2(x+4)-12\)
\(=[(x+2)(x+4)](x+3)^2-12\)
\(=(x^2+6x+8)(x^2+6x+9)-12\)
\(=a(a+1)-12\) (Đặt \(x^2+6x+8=a\) )
\(=a^2+a-12=(a-3)(a+4)=(x^2+6x+5)(x^2+6x+12)\)
\(=(x+1)(x+5)(x^2+6x+12)\)