đây không phải là lớp 1 bn ơi

1.=(x-2)(x 2+2x+7)+2(x-2)(x+2)-5(x-2) = 0
=>(x-2)(x 2+2x+7+2x+4-5) = 0
=>(x-2)(x 2+4x+6) = 0
Mà x 2+4x+6 (E Z)
=> x 2+4x+6 > 0
Vậy (x-2)=0 => x = 2
 

a) Ta có : n³ + 3n² + 2n

= n(n² + 3n + 2) 

= n(n + 1)(n + 2) \(⋮\)6 (tích 3 số nguyên liên tiếp) (đpcm)

b) A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + .... + 2⁹⁵ + 2⁹⁶ + 2⁹⁷ + 2⁹⁸ + 2⁹⁹

= (1 + 2 + 2² + 2³ + 2⁴) + 2⁵(1 + 2 + 2² + 2³ + 2⁴) + ... + 2⁹⁵(1 + 2 + 2² + 2³ + 2⁴)

= 31 + 2⁵.31 + .. + 2⁹⁵.31

= 31(1 + 2⁵ + ... + 2⁹⁵) \(⋮31\)(đpcm) 

c) Ta có 49ⁿ + 77ⁿ - 29ⁿ - 1

= (49ⁿ - 1) + (77ⁿ - 29ⁿ) 

= (49 - 1)(49^{n - 1} - 49^{n - 2} + .... - 1) + (77 - 29)(77^{n - 1} - 77^{n - 2}.29 + 77^{n- 3}.29² - .... - 1) 

= 48(49^{n - 1} - 49^{n - 2} + .... - 1) + 48(77^{n - 1} - 77^{n - 2}.29 + 77^{n- 3}.29² - .... - 1) 

= 48(49^{n - 1} - 49^{n - 2} + .... - 1 + 77^{n - 1} - 77^{n - 2}.29 + 77^{n- 3}.29² - .... - 1) \(⋮\)48 (đpcm) 

x = 1111 - 111 = 1000

x + 111 = 1111

x         = 1111 - 111

x         =     1000

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2: \(M=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{30}{3^{30}}\)

=>3M=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{30}{3^{29}}\)

=>3M+M=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{30}{3^{29}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{30}{3^{30}}\)

=>4M=\(1-\frac13+\frac{1}{3^2}-\cdots-\frac{1}{3^{29}}-\frac{30}{3^{30}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{29}}\)

=>3A=-1+\(\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{28}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{28}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{29}}\)

=>4A=\(-1-\frac{1}{3^{29}}=\frac{-3^{29}-1}{3^{29}}\)

=>\(A=\frac{-3^{29}-1}{4\cdot3^{29}}\)

Ta có: \(4M=1-\frac13+\frac{1}{3^2}-\cdots-\frac{1}{3^{29}}-\frac{30}{3^{30}}\)

\(=1+\frac{-3^{29}-1}{4\cdot3^{29}}-\frac{30}{3^{30}}=1+\frac{-3^{30}-3-120}{4\cdot3^{30}}=1-\frac14-\frac{123}{4\cdot3^{30}}=\frac34-\frac{123}{4\cdot3^{30}}\)

=>4M<3/4

=>M<3/16