40.1/3=40/3

40-0=40

3//8=3/8

ta có: \(A=\frac{3x+7}{x+1}=\frac{3x+3+4}{x+1}=\frac{3.\left(x+1\right)+4}{x+1}=\frac{3.\left(x+1\right)}{x+1}+\frac{4}{x+1}=3+\frac{4}{x+1}\)

Để A nguyên

\(\Rightarrow\frac{4}{x+1}\in z\)

\(\Rightarrow4⋮x+1\Rightarrow x+1\inƯ_{\left(4\right)}=\left(4;-4;2;-2;1;-1\right)\)

nếu x+1 = 4 => x = 3 (TM)

x +1 = -4 => x = -5 ( TM)

x + 1 = 2 => x = 1 ( TM)

x+ 1 = -2 => x = -3 ( TM)

x+1 = 1 => x = 0 ( TM)

x + 1 = -1 => x = -2 ( TM)

KL: x = .........

Để A nhận giá trị nguyên thì

\(\Leftrightarrow3x+7⋮x+1\)

\(\Leftrightarrow3\left(x+1\right)+4⋮x+1\)

\(\Leftrightarrow4⋮x+1\)

Vì \(x\inℤ\Rightarrow x+1\inℤ\)

\(\Rightarrow x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng giá trị

Đối chiếu điều kiện \(x\inℤ\)

Vậy \(x\in\left\{-2;-3;1;0;3;-5\right\}\)

\(1,25:\frac{15}{20}+\left(25\%-\frac{5}{6}\right):4.\frac{2}{3}\)

\(=\frac{5}{4}.\frac{20}{15}+\left(\frac{1}{4}-\frac{5}{6}\right).\frac{1}{4}.\frac{2}{3}\)

\(=\frac{5}{3}+\left(\frac{3}{12}-\frac{10}{12}\right).\frac{1}{6}\)

\(=\frac{5}{3}-\frac{7}{12}.\frac{1}{6}\)

\(=\frac{5}{3}-\frac{7}{72}=\frac{120}{72}-\frac{7}{72}=\frac{113}{72}\)

\(1,25:\frac{15}{20}+\left(25\%-\frac{5}{6}\right):4\cdot\frac{2}{3}\)

\(=\frac{5}{4}\cdot\frac{20}{15}+\left(\frac{1}{4}-\frac{5}{6}\right)\cdot\frac{1}{4}\cdot\frac{2}{3}\)

\(=\frac{5}{3}+\frac{-7}{12}\cdot\frac{1}{4}\cdot\frac{2}{3}\)

\(=\frac{5}{3}+\frac{-7}{48}\cdot\frac{2}{3}\)

\(=\frac{5}{3}+\frac{-7}{72}\)

\(=\frac{113}{72}\)

Khi tia Oy nằm giữa 2 tia Ox, Oz

\(\Rightarrow\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)

Chúc bạn học tốt !

Khi tia Oy nằm giữa 2 tia Ox và Oz

Tk mk nha!

Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

Ta có : 

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

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

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(............\)

\(\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow\)\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(\Rightarrow\)\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow\)\(A< 1+1-\frac{1}{50}< 1+1\)

\(\Rightarrow\)\(A< 2\)

Vậy \(A< 2\)

Chúc bạn học tốt ~ 

CM A < 2

=> CM \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\)

Ta thấy: \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{2}=\frac{1}{1.2}\)

             \(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{6}=\frac{1}{2.3}\)

Và cứ thế,....

             \(\frac{1}{50^2}=\frac{1}{2500}< \frac{1}{2450}=\frac{1}{49.50}\)

=>  \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

                                                                     \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

=>  \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\)

=> \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1^2}=1+1=2\)

=>ĐPCM

\(|3x+1|-17=-12\)

\(\Rightarrow|3x+1|=5\)

\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=4\\3x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}}\)

\(\left|3x+1\right|-17=-12\)

\(\Rightarrow\left|3x+1\right|=-12+17=5\)

\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=4\\3x=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\) ta có : 

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)

\(A=1-\frac{1}{2^{20}}\)

\(A=\frac{2^{20}-1}{2^{20}}\)

Vậy \(A=\frac{2^{20}-1}{2^{20}}\)

Chúc bạn học tốt ~ 

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

\(\Rightarrow2B-B=B=1-\frac{1}{2^{20}}< 1\)

\(\RightarrowĐPCM\)

\(M=1+2+2^2+...+2^{2013}\)

\(\Rightarrow2M=2+2^2+2^3+...+2^{2014}\)

\(\Rightarrow2M-M=2^{2014}-1\)

\(\Leftrightarrow M=2^{2014}-1\)

1/2 mũ 2 + 1/4 mũ 2+ 1/6 mũ 2 +....+1/1000 mũ 2