Lời giải:

$M=(x^{10}-24x^9)-(x^9-24x^8)+(x^8-24x^7)-(x^7-24x^6)+(x^6-24x^5)-(x^5-24x^4)+(x^4-24x^3)-(x^3-24x^2)+(x^2-24x)-(x-24)+1$

$=x^9(x-24)-x^8(x-24)+x^7(x-24)-.....+x(x-24)-(x-24)+1$

$=(x-24)(x^9-x^8+x^7-...+x-1)+1$

$=0.(x^9-x^8+....+x-1)+1=1$

\(M=x^{10}-25x^9+25x^8-25x^7+...-25x^3+25x^2-25x+25\)

Ta thấy : \(x=24\Rightarrow x+1=25\)

\(\Rightarrow M=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)

\(M=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(\Rightarrow M=1\)

Vậy \(M=1\left(tạix=24\right)\)

a) Biến đổi  x 2 – 2x + 1 = ( x   –   1 ) 2 ; thực hiện chia được kết quả x – 1.

b) Biến đổi 8 x 3  + 27 = (2x + 3)(4 x 2  – 6x + 9); thực hiện phép chia được kết quả 4 x 2  – 6x + 9.

c) Phân thích x 6   –   6 x 4  + 12 x 2  – 8 = ( x 2 – 2)( x 4  – 4 x 2  + 4); thực hiện phép chia được kết quả - x 4  + 4 x 2  – 4.

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

f ) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+5=t\), ta có :

\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)

Thay và ta có :

\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

25 x 3 = 25 x 2 . x = 5 x x = 5 x x ( với x > 0 )

Ta có: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3-3x^2+16x^2-8x-6x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[3x^2\left(2x-1\right)+8x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+8x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+9x-x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left[3x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-1=0\\x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\2x=1\\x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\\x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{1}{2};-3;\dfrac{1}{3}\right\}\)

Ta có: f"(x) = 2x - 1

Do đó phương trình vô nghiệm.