13x 12=
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x10 - 13x9 + 13x8 - ... - 13x + 13
= (x10 - 12x9) + (- x9 + 12x8) + ... + (- x + 12) + 1
= x9(x - 12) + x8(- x + 12) +...+ (- x + 12) + 1 = 1
S(x) = x^9(x - 12) -x^8(x - 12) + x^7(x - 12) + . . . +x(x-12) - (x - 12) - 2
Suy ra: S(x) = -2
g(x) = x14 - 13x13 + 13x12 - 13x11 + ... + 13x2 - 13x + 15
= x14 - (12 + 1)x13 + (12 + 1)x12 - (12 + 1)x11 + ... + (12 + 1)x2 - (12 + 1)x + 15
Tại x = 12 thì ta có:
g(12) = x14 - (x + 1)x13 + (x + 1)x12 - (x + 1)x11 + ... + (x + 1)x2 - (x + 1)x + 15
= x14 - x14 - x13 + x13 + x12 - x12 - x11 + ... + x3 + x2 - x2 - x + 15
= -x + 15
Thay x = 12, ta có:
g(12) = -12 + 15 = 3
Vậy g(12) = 3
\(x^4-13x^3+13x^2-13x+2014\)
\(=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+2002\)
\(-x^4-x^4-x^3+x^3+x^2-x^2-x+x+2002\)
\(=2002\)
\(x^4-13x^3+13x^2-13x+2014\)
\(=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+2002\)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+x+2002\)
\(=2002\)
x = 79 => 80 = x + 1 thay vòa Px ta có
P(x) = x^7 - ( x + 1) x^6 + .... + (x + 1) x + 15
= x^7 - x^7- x^6 + ... + x^2 + x +15
= x + 15
= 79 + 15
= 94
Ý B tương tự
Ta có: \(x^3-12=13x\)
\(\Leftrightarrow x^3-13x-12=0\)
\(\Leftrightarrow x^3-x-12x-12=0\)
\(\Leftrightarrow x\left(x^2-1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-12\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+3x-12\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\\x=-3\end{matrix}\right.\)
Vậy: S={-1;4;-3}
a \(\Leftrightarrow3x^2+9x+4x+12=0\Leftrightarrow3x\left(x+3\right)+4\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(3x+4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(3x^2+13x+12=0\)
\(\Leftrightarrow3\left(x^2+\dfrac{13}{3}x+4\right)=0\Leftrightarrow x^2+\dfrac{13}{3}x+4=0\)
\(\Leftrightarrow x^2+3x+\dfrac{4}{3}x+4=0\)
\(\Leftrightarrow x\left(x+3\right)+\dfrac{4}{3}\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{4}{3}\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=-3\end{matrix}\right.\)
13 x 12 = 156 nhé
k ch mk nha
156 nhá