đề bài

2x²-4x+2

=2(x²-2x+1)

=2(x-1)²

x²-9+(x-3)

= (x²-3²)+(x-3)

=(x-3)(x+3)+(x-3)

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

=(x-3)(x+4)

\(x^2-9+\left(x-3\right)=x^2+x-12=x^2-3x+4x-12=x\left(x-3\right)+4\left(x-3\right)=\left(x-3\right)\left(x+4\right)\)

\(A=11...1\left(2n\right);B=11...1\left(n+1\right);C=66...6\left(n\right)\)

\(\Rightarrow A+B+C+8=11...1\left(2n\right)+11...1\left(n+1\right)+66...6\left(n\right)+8\)

\(=11...1\left(n\right).10^n+11...1\left(n\right)+11...1\left(n\right).10+1+6.11...1\left(n\right)+8\)

\(=11...1\left(n\right).10^n+17.11...1\left(n\right)+9\)

Đặt\(11...1\left(n\right)=a\)

\(\Rightarrow10^n=9a+1\)

\(\Rightarrow A+B+C+8=a\left(9a+1\right)+17a+9\)

\(=9a^2+18a+9a=\left(3a+3\right)^2\)

Thay \(a=11...1\left(n\right)\Rightarrow A+B+C+8=\left(3.11...1\left(n\right)+3\right)^2\)

Chú thích: n;n+1;2n là số chữ số

PTHH là:

 2Fe(OH)_{3 ===>}Fe₂O₃+3H₂O

nhầm phải là: 2-0-3

\(\left(x-2+3y\right)\left(x+3y+2\right)\)

\(=\left(x+3y-2\right)\left(x+3y+2\right)\)

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

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

(x+3y)\(^2\)- 4

Theo đầu bài ta có:
\(49^5-49\)
\(\Leftrightarrow\left[\left(49^2\right)^2\cdot49\right]-49\)
\(\Leftrightarrow\left[\left(...01\right)^2\cdot49\right]-49\)
\(\Leftrightarrow\left[\left(...01\right)\cdot49\right]-49\)
\(\Leftrightarrow\left(...49\right)-49\)
\(\Leftrightarrow\left(...00\right)\)
Do 49⁵ - 49 có 2 chữ số 0 ở cuối nên chia hết cho 100. Vậy 49⁵ - 49 chia hết cho 100        ( đpcm )

Theo đầu bài ta có:
\(\left(a-b\right)\left(a+b\right)+3\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a-b+3\right)\)

=a²-b²+3a+3b

=aa-bb+3a+3b

=a(a+3)-b(b-3)