\(A=7+7+7^2+...+7^{100}\)

\(7A=7^2+7^2+7^3+...+7^{101}\)

\(A=14+7^2+7^{101}\)

Em xem thử lại đề bài nhé

\(A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8\)

\(A=\left(7+7^3\right)+\left(7^2+7^4\right)+\left(7^5+7^7\right)+\left(7^6+7^8\right)\)

\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)

\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)

\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)

\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)

Ta có: 5 ⋮ 5

⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm) 

A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78

A =  (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)

A = 7.(1 + 72)  + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)

A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)

A = 7.50 + 72.50 + 75.40 + 76.50

A = 50.(7 + 72 + 75 + 76)

Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm

A = 7 + 7² + 7³ + 7⁴ + 7⁵ + 7⁶ + 7⁷ + 7⁸

A =  (7 + 7³) + (7²+ 7⁴) + (7⁵ + 7⁷) + (7⁶ + 7⁸)

A = 7.(1 + 7²)  + 7².(1 + 7²) + 7⁵.(1 + 7²) + 7⁶.(1 + 7²)

A = 7.( 1 + 49) + 7².( 1 + 49) + 7⁵.(1 + 49) + 7⁶. (1 + 49)

A = 7.50 + 7².50 + 7⁵.50 + 7⁶.50

A = 50.(7 + 7² + 7⁵ + 7⁶)

Vì 50 ⋮ 5 nên A = 50.(7 + 7² + 7⁶) ⋮ 5 đpcm

1. x²-x-2       

      =(x²-2x)+(x-2)

       = x(x-2)+(x-2) 

       = (x+1)(x-2)

2.x²-3x+2

=x²-x-2x+2

=(x²-x)-(2x-2)

=x(x-1)-2(x-1)

=(x-2)(x-1)

3.-x²-2x+3

=3-2x-x²

=3+x-3x-x²

=(3+x)-(3x+x²)

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

=(1-x)(3+x)

4. x²-5x+4

=x²-x-4x+4

=(x²-x)-(4x-4)

=x(x-1)-4(x-1)

=(x-1)(x-4)

5. x²-5x+6

=x²-2x-3x+6

=(x²-2x)-(3x-6)

=x(x-2)-3(x-2)

=(x-2)(x-3)

6.x²-6x+5

=(x²-x)-(5x-5)

=x(x-1)-5(x-1)

=(x-1)(x-5)

7.x²-7x+12

=(x²-3x)-(4x-12)

=x(x-3)-4(x-3)

=(x-4)(x-3)

8.-x²+7x-12

=(-x²+3x)+(4x-12)

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

=(4-x)(x-3)

9.x²-3x-4

=(x²+x)-(4x+4)

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

=(x-4)(x+1)

mik làm 1 nửa thôi dài quá

bạn đăng nhỏ câu hỏi ra

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

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

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

4) \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

5) \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

6) \(x^2-6x+5=\left(x-1\right)\left(x-5\right)\)

\(7^1+7^2+7^3+...+7^{117}+7^{118}=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{116}\left(1+7+7^2\right)\)

\(=7.57+7^4.57+...+7^{116}.57=57\left(7+7^4+...+7^{116}\right)⋮57\)

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

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