1. A = \(\dfrac{3n-7}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{-7}{n-1}=3+\dfrac{-7}{n-1}\)

Tại giá trị \(A\notin Z,3\in Z\)\(\Rightarrow\dfrac{-7}{n-1}\in Z\)\(\Rightarrow n-1\inƯ\left(-7\right)\) với \(x\ne1\) (mẫu sẽ có giá trị là 0 nếu x = 1)

Tại \(n-1=7\)\(\Leftrightarrow n=7+1=8\)

Tại \(n-1=-7\Leftrightarrow n=-7+1=-6\)

Tại \(n-1=1\Leftrightarrow n=1+1=2\)

Tại \(n-1=-1\Leftrightarrow n=-1+1=0\)

2. B = \(\dfrac{4n+1}{2n-3}=\dfrac{4n+6}{2n-3}+\dfrac{-5}{2n-3}=2+\dfrac{-5}{2n-3}\)

Tại giá trị \(B\in Z,2\in Z\)\(\Rightarrow\dfrac{-5}{2n-3}\in Z\)\(\Rightarrow2n-3\inƯ\left(-5\right)\) với \(x\ne\dfrac{3}{2}\)

Tại \(2n-3=5\Leftrightarrow2n=8\Leftrightarrow n=4\)

Tại \(2n-3=-5\Leftrightarrow2n=-2\Leftrightarrow n=-1\)

Tại \(2n-3=1\Leftrightarrow2n=4\Leftrightarrow n=2\)

Tại \(2n-3=-1\Leftrightarrow2n=2\Leftrightarrow n=1\)

\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\Rightarrow4S=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}+......-\dfrac{1}{2^{2002}}\Rightarrow4S+S=5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< 0,2\left(\text{đpcm}\right)\)

\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)

\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)

đề bài là gì

a)

\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)

\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

..........................

\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)

\(\Leftrightarrow A< \dfrac{2003}{2004}\)

\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)

\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)

b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)

\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)

\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)

\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)

\(\Leftrightarrow S< 0,2\left(đpcm\right)\)

cho mik hỏi mik ko hiểu tại sao từ 1/2^4n-2 khi nhân với 2^2 lại ra đc 1/2^4n vậy? Xin hãy giải đáp giùm mik

a) ( x + 5 )³ = -64

x + 5 = - 4

x = - 4 - 5

x = -9

b) (2x - 3)²=9

2x - 3 = 3

2x = 3+3

2x = 6

x = 6 : 2

x = 3

e) \(\dfrac{8}{2x}=4\)

=> 4 . 2x = 8

8x =8

x = 8 : 8

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)

=> x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(\dfrac{1}{4}.x=\dfrac{1}{32}\)

x = \(\dfrac{1}{32}:\dfrac{1}{4}\)

x = \(\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\dfrac{-1}{27}\)

a) (x + 5)³ = -64

=> (x + 5)³ = (-4)³

x + 5 = -4

x = -4 - 5

x = -9

b) (2x - 3)² = 9

=> (2x - 3)² = (\(\pm\)3)²

=> 2x - 3 = 3 hoặc 2x - 3 = -3

*2x - 3 = 3

2x = 3 + 3

2x = 9

x = \(\dfrac{9}{2}\)

*2x - 3 = -3

2x = -3 + 3

2x = 0

x = 0 : 2

x = 0

Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)

c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)

\(\dfrac{x}{2}=8\)

x = 8 : 2

x = 4

d) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

=> (-2)ⁿ . (-2)²= (-2)⁵

(-2)ⁿ = (-2)⁵ : (-2)²

(-2)ⁿ = (-2)³

Vậy n = 3

e) \(\dfrac{8}{2x}=4\)

=> 2x . 4 = 8

2x = 8 : 4

2x = 2

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)

2x - 1 = 3

2x = 3 + 1

2x = 4

x = 4 : 2

x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)

\(x=\left(\dfrac{1}{2}\right)^3\)

\(x=\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^3\)

\(x=\dfrac{-1}{27}\).

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

Bài 1:

1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=3^2=9\)

2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)

\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)

\(=2^7:2^3:\dfrac{1}{2^4}\)

\(=2^4.2^4=256\)

3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)

\(=\dfrac{43}{48}\)

4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=-3-1+\dfrac{1}{8}\)

\(=-4+\dfrac{1}{8}\\ \)

\(=-\dfrac{31}{8}\)

5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)

Chúc bạn học tốt

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\dfrac{1}{4}:2\)

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

A=\(\dfrac{7^2-1}{7^4}+\dfrac{7^2-1}{7^8}+...+\dfrac{7^2-1}{7^{100}}=\left(7^2-1\right)\left(\dfrac{1}{7^4}+\dfrac{1}{7^8}+...+\dfrac{1}{7^{100}}\right)=48\cdot B\)Dễ dàng tính được B( nhân hết với 7 mũ 4 roi trừ đi, chia ra là xong) ra đpcm.

Lên lớp 11 thì ta có dạng tổng quát luôn này(tức là nếu n quá lớn thì có thể coi là xảy ra dấu bằng) \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^n}-\dfrac{1}{7^{n+2}}< \dfrac{1}{50}\)