1. A = 75(4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 25

    A = 25 . [3 . (4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 1]

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 3 + 1)

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 4)

    A = 25 . 4 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1) \(⋮\) 100

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

\(A=\frac{3}{4}-1+1=\frac{3}{4}\)

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

2.

A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\

A=0

Bài 3:

a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)

2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)

2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)

3A = \(1-\frac{1}{2^6}\)

=> 3A < 1 

=> A < \(\frac{1}{3}\)(đpcm)

b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)

4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)       (1)

Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)

4B = \(3-\frac{1}{3^{99}}\)

=> 4B < 3

=> B < \(\frac{3}{4}\)   (2)

Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)

bài 1:

5n+7 chia hết cho 3n+2

=> [3(5n+7) - 5(3n + 2)] chia hết cho 3n+2

=> (15n + 21 - 15n - 10) chia hết cho 3n+2

=> 11 chia hết cho 3n + 2

=> 3n + 2 thuộc Ư(11) = {1;-1;11;-11}

Ta có bảng:

Vậy n = {-1;3}

Bài 2:

1, chữ số tận cùng

a, Xét 7¹⁹⁹⁹

Ta có: 7¹⁹⁹⁹ = 7¹⁹⁹⁶.7³ = (7⁴)⁴⁹⁹.343 = (...1)⁴⁹⁹.343 = (....1).343 = ....3 (1)

Vậy số 57¹⁹⁹⁹ có tận cùng là 3

b, Xét 3¹⁹⁹⁹

Ta có: 3¹⁹⁹⁹ = 3¹⁹⁹⁶.3³ = (3⁴)⁴⁹⁹.27 = (...1)⁴⁹⁹.27 = (...1) . 27 = ....7  (2)

Vậy số 93¹⁹⁹⁹ có chữ số tận cùng là 7

2, 

Từ (1) và (2) suy ra A = 999993¹⁹⁹⁹ + 555557¹⁹⁹⁹ = ...7 + ...3 = ....0

Vì A có chữ số tận cùng là 0 nên A chia hết cho 5. 

Câu 8( Mình không viết đè nữa nha)

a)   2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100

=  1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100

=  1 – 1/100 < 1

=   99/100 < 1

    Vậy A< 1

2.

a) Ta có:

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)

Vậy x = -1

b) Ta có:

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)

Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)

Vậy, x = -2004

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

Bước cưối 58/21 minh man viết nhầm nên sai 

\(\left(x-\frac{9}{4}\right)=\frac{58}{21}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{87}{7}\)

\(x=\frac{87}{7}+\frac{9}{4}\)

\(x=\frac{411}{28}\)

b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)

\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)

Suy ra:

\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)

\(\Leftrightarrow\)2x-x²-x+2+5x²-5 = 15x-15

\(\Leftrightarrow\)2x-x²-x+5x²-15x = -15+5-2

\(\Leftrightarrow\)4x²-14x = -12

\(\Leftrightarrow4x^2-14x+12=0\)

\(\Leftrightarrow4x^2-8x-6x+12=0\)

\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0

\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)

Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)

câu 1b

Gọi d là ƯCLN (3n-7, 2n-5), d thuộc N*

Ta có : 3n-7 chia ht cho d , 2n_5 chia ht cho d

suy ra: 2(3n-7) chia ht cho d ,  3(2n-5) chia ht cho d

suy ra 6n-14 chia ht cho d, 6n-15 chia ht cho d

dấu suy ra [(6n -15) - (6n-14)] chia ht cho d dấu suy ra 1 chia ht cho d suy ra d =1

Vậy......

1) b. Để chứng tỏ \(\frac{3n-7}{2n-5}\) là phân số tối giản 

Ta cần chứng minh: ( 3n - 7; 2n - 5 ) = 1 

Thật vậy: ( 3n - 7 ; 2n - 5 ) = ( 2n - 5 ; ( 3n - 7 ) - ( 2n - 5 ) )  = ( 2n - 5; n - 2 ) = ( n - 2; n - 3 ) = ( n - 2; 1 ) = 1

=> \(\frac{3n-7}{2n-5}\) là phân số tối giản 

3) \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{12}\)

Ta có: \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\left(\frac{1}{5}+\frac{1}{7}\right)+\frac{1}{6}=\frac{12}{35}+\frac{1}{6}>\frac{12}{36}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{1}{2}\)

\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}=\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}\right)>\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \)

=> A > 1/2 + 1/2 + 1/2 + 1/2 = 2