Ta có:

\(P\left(9\right)-P\left(6\right)=2019\)

\(\Leftrightarrow81a+9b+c-36a-6b-c=2019\)

\(\Leftrightarrow45a+3b=2019\)

Lại có:

\(P\left(10\right)-P\left(7\right)\)

\(=100a+10b+c-49a-7b-c\)

\(=51a+3b\)

\(=\left(45a+3b\right)+6a\)

\(=2019+6a\) là số lẻ vì  \(6a\) là số chẵn và \(2019\) lẻ

=> ĐPCM

P/S:Hiện tại chỉ nghĩ ra bài 2

\(\frac{\left(2n+5\right)^2-25}{4}=\frac{\left(2n+5-5\right)\left(2n+5+5\right)}{4}=\frac{2n\left(2n+10\right)}{4}=\frac{4n\left(n+5\right)}{4}\)

=> điều phải chứng minh!

(2n+5)^2-25

=(2n+5)^2-5^2

=(2n+5-5)(2n+5+5)

=2n(2n+10)

=4n^2+20n

Có n nguyên=> 4n^2,20n nguyên=>4n^2 và 20n chia hết cho 4=>điều phải chứng minh

\(\left(3x-2\right)\left(x^2-4x+5\right)\)

\(=3x^3-12x^2+15x-2x^2+8x-10\)

\(=3x^3-14x^2+23x-10\)

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

\(2x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(2x-3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)

D = x - x² + 3

D = - x² + x + 3

D = - ( x² - x - 3 )

D = - [ x² - 2 . x . 1 / 2 + ( 1 / 2 )² - ( 1 / 2 )² - 3 ]

D = - [ ( x - 1 / 2 )² - 13 / 4 ]

D = - ( x - 1 / 2 )² + 13 / 4 \(\le\)13 / 4

Dấu " = " xảy ra \(\Leftrightarrow\)x - 1 / 2 = 0

                             \(\Rightarrow\)x              = 1 / 2

Max D = 13 / 4 \(\Leftrightarrow\)x = 1 / 2

D=x-x^2+3

D= -[x^2 -x +1/4 ] + 13/4 

D=-(x-1/2)^2 +13/4 

Vì -(x-1/2)^2<=0 => D<=13/4

Dấu = xảy ra <=> x-1/2=0 <=> x=1/2

\(b.=4\left(x^2+4x+4\right)\)

\(=4\left(x+2\right)^2\)

Học tốt

\(x^3-2x^2+x-xy^2\)

\(=x\left(x^2-2x+1-y^2\right)\)

\(=x\left[\left(x-1\right)^2-y^2\right]\)

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

Câu a ko rõ đề bài lắm

\(b.\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\left(x+2\right).4=0\)

\(\Rightarrow x+2=0\)

\(x=-2\)