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

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

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

\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)

\(=5^{2021-\dfrac{1}{2021}}\)

\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)

\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)

Bạn ơi thiếu đề rồi, cái biểu thức này không tính được đâu , mình nghĩ thế

m = 5 

n = -1

mình nhầm câu trên

\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)

\(n=-36\)

Ta có:  \(2^n=\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}....\dfrac{30}{62}.\dfrac{31}{64}\)

⇔ \(2^n=\dfrac{1.2.3.4....31}{2.\left(2.3.4.....1\right).64}=\)

⇔ \(2^n=\dfrac{1}{2}.\dfrac{1}{64}=\dfrac{1}{128}\)  \(\Leftrightarrow\)   \(2^n=\dfrac{1}{2^6}\)

⇔ \(2^{x+6}=1\)

⇔ \(x+6=0\)

⇒  \(\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Để phân số nguyên thì n + 10 chia hết cho 2n - 8

=> 2.(n + 10) chia hết cho 2n - 8

=> 2n + 20 chia hết cho 2n - 8

=> 2n - 8 + 28 chia hết cho 2n - 8

Do 2n - 8 chia hết cho 2n - 8 => 28 chia hết cho 2n - 8

Do  n ∈ N⇒2n − 8 ≥ −8 mà 2n - 8 là số chẵn

=> 2n − 8 ∈ { −2;2; − 4;4;14;28 }

=> 2n ∈  { 6;10;4;12;22;36 }

=> n ∈ { 3;5;2;6;11;18 }

các bn chỉ cần lm phần phân số tối giản thôi còn giá trị số nguyên mk lm đc rồi

c)\(7^{2n}+7^{2n+2}=2450\)

⇒\(7^{2n}+7^{2n}.7^2=2450\)

⇒\(7^{2n}.50=2450\)

⇒\(7^{2n}=49\)\(=7^2\)

⇒2n=2

⇒n=1

a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\)                   b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)

\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\)                    \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)

⇒n=3                                          ⇒m=2

1/

a/ A = 1 + 3 + 3^2 + 3^3 + ... + 3^119

=> 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^120

=> 3A - A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^120 - (1 + 3 + 3^2 + 3^3 + ... + 3^119)

=> 2A = 3^120 - 1

=> A = (3 ^120 - 1)/2

b/ 2A + 1 = 27^x

<=> 3^120 = 27^x

^<=> 27^40 = 27^x

<=> x = 40

c/ +) A = 1 + 3 + 3^2 + 3^3 + ... + 3^119

= (1 + 3^2) + (3 + 3^3) + (3^4 + 3^6) + ...+ (3^117 + 3^119)

= 1+ 3^2 + 3(1+ 3^2) + 3^4(1 + 3^2) ...+ 3^117( 1+ 3^2)

= (1 + 3^2) (1 + 3 + 3^4+ ...+ 3^117)

= 10 * (1 + 3 + 3^4+ ...+ 3^117) \(⋮\) 5

+) A = 1 + 3 + 3^2 + 3^3 + ... + 3^119

= (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ...+ (3^117 + 3^118 + 3^119)

= (1 + 3 + 3^2) + 3^3 (1+ 3 + 3^2) + ...+ 3^117 (1+ 3 + 3^2)

= (1 + 3 + 3^2) (1+ 3^3 +... + 3^117)

= 13 * (1+ 3^3 +... + 3^117) \(⋮\)13

2b

Câu hỏi của Raf - Toán lớp 6 - Học toán với OnlineMath

3/ \(\dfrac{x-3}{3}=\dfrac{7}{y+2}\)

đk: y # - 2

<=> (x - 3) (y + 2) = 21

=> x - 3 = 7 ; y + 2 = 3 <=> x = 10; y = 1

hoặc x - 3 = -7; y + 2 = -3 <=> x = -4. y = -5

hoặc x - 3 = 3; y + 2 = 7 <=> x = 6; y = 5

hoặc x- 3 = -3; y + 2 = -7 <=> x = 0; y = -9

hoặc x - 3 = 1; y + 2 = 21 <=> x = 4; y = 19

hoặc x - 3 = -1; y + 2 = -21 <=> x = 2; y = -23

hoặc x - 3 = 21; y + 2 = 1 <=> x = 24; y = -1

hoặc x - 3 = -21; y + 2 = -1 <=> x = -18; y = -3

vậy (x;y) = (10;1) ; (-4;-5) ; (6;5) ; (0,-9); (4;19); (2;-23); (24;-1); (-18;-3)

Câu 1:

\(A\in Z\Rightarrow6n-1⋮3n+2\)

\(\Rightarrow6n+4-5⋮3n+2\)

\(\Rightarrow2\left(3n+2\right)-5⋮3n+2\)

\(\Rightarrow5⋮3n+2\)

đến đây tự lm nốt nhé

1. Để A có giá trị nguyên thì \(6n-1⋮3n+2\)

Ta có: \(\left\{{}\begin{matrix}6n-1⋮3n+2\\3n+2⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\2\left(3n+2\right)⋮3n+2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n+4⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n-1+5⋮3n+2\end{matrix}\right.\)

\(\Rightarrow\left(6n-1+5\right)-\left(6n-1\right)⋮3n+2\)

\(\Rightarrow5⋮3n+2\)

\(\Rightarrow3n+2\inƯ\left(5\right)\)

\(\Rightarrow3n+2\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow3n\in\left\{-7;\pm3;-1;\right\}\)

\(\Rightarrow n\in\left\{\pm1\right\}\)

Vậy để \(A\in Z\) thì n nhận các giá trị là: \(\pm1\)

2. Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

Ta có: \(2B=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2B=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(2B-B=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(B=\dfrac{1}{2}-\dfrac{1}{2^{100}}\)

\(\Rightarrow B< \dfrac{1}{2}< \dfrac{2}{2}=1\)

\(\Rightarrow B< 1\left(đpcm\right)\)