\(=\dfrac{3x^3-3x^2-21x-21+12}{3x-3}\)

\(=x^2-7+\dfrac{4}{x-1}\)

Chia đa thức 21x – 4 cho 3x² được thương là 0, dư 21x – 4

Đáp án là B

Ta có y ' = 3 x 2 + 6 x − 21  .

Hàm số có 2 cực trị  x 1 ; x 2 ⇒ x 1 . x 2 = c a = − 7.

=18x^2-3x+24x-4

=3x(6x-1)+4(6x-1)

=(6x-1)(3x+4)

18x² + 21x - 4

= 18x² +24x - 3x - 4

= 6x(3x + 4) - (3x + 4)

= (6x - 1)(3x + 4)

^{A = x^100 - 21x^99 - 21x^98 - 21x^97 -...-21x^2 - 21x +2010}

^{A=x^100 - 22x^99 + x^99 -22x^98 + x^98 - ... - 22x +x +2010}

^{A=x^99 (x-22) + x^98 (x-22) + x^97(x-22) + ... + x(x-22) + x +2010}

^{A=(x-22) (x^99 + x^98 + x^97 + ... + x) + x + 2010}

Thay x = 22 vào A, tao có:

A= (22-22) (22^99 + 22^98 + ... +22) + 22 + 2010

A = 0 (22^99 + 22^98 + ... +22) + 2032

A= 0 + 2032

A = 2032

x=22

=>x-1=21

thay 21=x-1 vào A ta được:

A=x¹⁰⁰-(x-1)x⁹⁹-(x-1)x⁹⁸-(x-1)x⁹⁷-...-(x-1)x²-(x-1)x+2010

=x¹⁰⁰-x¹⁰⁰+x⁹⁹-x⁹⁹+x⁹⁸-x⁹⁸+x⁹⁷-...-x³+x²-x²+x+2012

=>A=x+2012

thay x=22 vào A=x+2012 ta được:

A=22+2012=2034

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Bài 1

a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)

b/

\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)

\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

1.

c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)

\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)

d/

\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)