cái này là toán mà bạn, đâu phải vật lý

\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)

<=> \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}=\frac{56}{673}\)

<=> \(4.\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}\right)=4.\frac{56}{673}\)

<=> \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}=\frac{224}{673}\)

<=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{224}{673}\)

<=> \(\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)

<=> \(\frac{n+4-3}{3.\left(n+4\right)}=\frac{224}{673}\Leftrightarrow\frac{n}{3.\left(n+4\right)}=\frac{224}{673}\)

<=> 673n = 224.3(n+4)

<=> 673n = 224.3.n + 224.3.4

<=> 673n = 672n + 2688

<=> 673n - 672n = 2688

<=> n = 2688

\(A=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}+...+\frac{1}{n^2}-\frac{1}{4n}=\frac{56}{673}\)

\(\Rightarrow4A=\)

\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)

\(\Rightarrow\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n\left(n+4\right)}=\frac{56}{673}\)

\(\Rightarrow\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}\right)=\frac{56}{673}\)

\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+4}\right)=\frac{56}{673}\)

\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)=\frac{56}{673}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{56}{673}:\frac{1}{4}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)

\(\Rightarrow\frac{1}{n+4}=\frac{1}{3}-\frac{224}{673}\)

\(\Rightarrow\frac{1}{n+4}=\frac{1}{2019}\)

=> n + 4 = 2019 

     n = 2019 - 4

     n = 2015

ta có 

1/1*2+1/2*3+1/3*4+...+1/n*(n+1)=1/1-1/2+1/2-1/3+1/3-...-1/n+1= 33/34 (quy tắc)

                                                    1 - 1/n+1=33/34

                                                     1/n+1=1/34  

                                                     nên n =33

\(\frac{x-1}{9}=\frac{8}{3}\Rightarrow\)\(\frac{x-1}{9}=\frac{24}{9}\Rightarrow x-1=24\)

                                        x=24+1

                                        x=25

Vậy x=25

\(\frac{x-1}{9}=\frac{8}{3}\)

\(\Leftrightarrow\left(x-1\right):9=\frac{8}{3}\)

\(\Leftrightarrow\left(x-1\right)=24\)

\(\Leftrightarrow x=24+1\)

\(\Leftrightarrow x=25\)

\(\frac{x-1}{9}\)=\(\frac{8}{3}\)

\(\Rightarrow\)\(\frac{x-1}{9}\)=\(\frac{24}{9}\)

\(\Rightarrow\)x-1=24

Vậy x =25

1) \(-x-3=-2\left(x+7\right)\\ \Rightarrow-x-3=-2x-14\\ \Rightarrow-x+2x=-14+3\\ \Rightarrow x=-11\)

2) \(A=\frac{12}{\left(x+1\right)^2+3}\\ Tac\text{ó}:\left(x+1\right)^2\ge0\\ \Rightarrow\left(x+1\right)^2+3\ge3\\ \Rightarrow A\le\frac{12}{3}=4\)

Max A=4 khi x=-1

3) Đăt : \(n^2+4=k^2\\ \Rightarrow k^2-n^2=4\\ \Rightarrow\left(k-n\right)\left(k+n\right)=4\)

lập bang ra rồi tính

Bài 2: 

a: Để E là số nguyên thì \(3n+5⋮n+7\)

\(\Leftrightarrow3n+21-16⋮n+7\)

\(\Leftrightarrow n+7\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

hay \(n\in\left\{-6;-8;-5;-9;-3;-11;1;-15;9;-23\right\}\)

b: Để F là số nguyên thì \(2n+9⋮n-5\)

\(\Leftrightarrow2n-10+19⋮n-5\)

\(\Leftrightarrow n-5\in\left\{1;-1;19;-19\right\}\)

hay \(n\in\left\{6;4;29;-14\right\}\)

gọi UCLN ( 14n+ 3 ; 21n +5 ) là d

=> 14n+ 3⋮d và 21n +5⋮d

=> 42n + 9⋮d và 42n + 10⋮d

=> 42n + 10 - (42n + 9) ⋮ d

=> 42n + 10 - 42n - 9⋮ d

=> 1⋮ d

=> p/s ...là phân số tối giản

1) Để phân số \(\frac{14n+3}{21n+5}\) là PSTG thì

ƯC(14n+3, 21n+5)={-1,1}

Gọi d là UC của 14n+3 và 21n+5

⇒14n+3⋮d

21n+5⋮d

⇒3(14n+3)⋮d

2(21n+5)⋮d

⇒42n+9⋮d

42n+10⋮d

⇒42n+9-(42n+10)⋮d

⇒42n+9-42n-10⋮d

⇒-1⋮d

⇒d={1, -1)

⇒ƯC(14n+3, 21n+5)={-1,1}

Vậy phân số................

2)\(\text({\frac{1}{4}.x+\frac{3}{4}.x})^{2}\)=\(\frac{5}{6}\)

⇒\(\text((\frac{1}{4}+\frac{3}{4}).x)^2=\frac{5}{6}\)

⇒\(\text{(1x)}^2\)=\(\frac{5}{6}\)

⇒x=....(mình ko tính dc)

Vậy x∈ϕ

3) A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)

=\(\frac{3.8.15...899}{4.9.16...900}\)

=\(\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)

=\(\frac{1.2.3...29}{2.3.4...30}.\frac{3.4.5....31}{2.3.4...30}\)

=\(\frac{1}{30}.\frac{31}{2}\)

=\(\frac{31}{60}\)