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\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)
=> \(1+\frac{1}{\frac{2\left(1+2\right)}{2}}+\frac{1}{\frac{3\left(1+3\right)}{2}}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=2\)
=> \(1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\)
=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}\)
=> \(\frac{1}{x+1}=0\Rightarrow x\in\varnothing\)
Bài làm :
Ta có :
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)
\(\Leftrightarrow1+\frac{1}{\frac{2\left(1+2\right)}{2}}+\frac{1}{\frac{3\left(1+3\right)}{2}}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=2\)
\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}=0\)
=> Không tồn tại x
1 Giải :
\(\frac{3x+7}{x-1}\)là phân số <=> x - 1 \(\ne\)0 => x \(\ne\)1
Ta có : \(\frac{3x+7}{x-1}=\frac{3\left(x-1\right)+8}{x-1}=3+\frac{8}{x-1}\)
Để \(\frac{3x+7}{x-1}\)là số nguyên thì 8 \(⋮\)x - 1 => x - 1 \(\in\)Ư(1; -1; 2; -2; 4; -4; 8; -8}
Lập bảng :
| x - 1 | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 |
| x | 2 | 0 | 3 | -1 | 5 | -3 | 9 | -7 |
Vậy x \(\in\){2; 0; 3; -1; 5; -3; 9; -7} thì \(\frac{3x+7}{x-1}\)là số nguyên
Đặt \(A=\frac{3x+7}{x-1}\)
Ta có: \(A=\frac{3x+7}{x-1}=\frac{3x-3+10}{x-1}=\frac{3x-3}{x-1}+\frac{10}{x-1}=3+\frac{10}{x-1}\)
Để \(A\in Z\)thì \(\frac{10}{x-1}\in Z\Rightarrow10⋮x-1\Leftrightarrow x-1\in U\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(5\) | \(-5\) | \(10\) | \(-10\) |
| \(x\) | \(2\) | \(0\) | \(3\) | \(-1\) | \(6\) | \(-4\) | \(11\) | \(-9\) |
Vậy, với \(x\in\left\{-9;-4;-1;0;2;3;6;11\right\}\)thì \(A=\frac{3x+7}{x-1}\in Z\)
1.Ta có: \(\frac{x}{3}=-\frac{12}{9}\)
=> \(\frac{3x}{9}=-\frac{12}{9}\)
=> 3x = -12
=> x = -12 : 3
=> x = -4
\(\frac{4}{5}x-\frac{8}{5}=-\frac{1}{2}\)
=> \(\frac{4}{5}x=-\frac{1}{2}+\frac{8}{5}\)
=> \(\frac{4}{5}x=\frac{11}{10}\)
=> \(x=\frac{11}{10}:\frac{4}{5}\)
=> \(x=\frac{11}{8}\)
1. \(\frac{-7}{12}\)< \(\frac{x-1}{4}\)< \(\frac{2}{3}\)
=> \(\frac{-7}{12}\)< \(\frac{3.\left(x-1\right)}{12}\)< \(\frac{8}{12}\)
=> 3 . ( x - 1 ) thuộc { - 6 ; - 5 ; - 4 ; - 3 ; - 2 ; - 1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7}
Lập bảng tính giá trị x , cái này dễ lên bạn tự làm nha
1/ \(-\frac{7}{12}< \frac{x-1}{4}< \frac{2}{3}\)
hay \(\frac{-7}{12}< \frac{3.\left(x-1\right)}{12}< \frac{8}{12}\)
Vậy \(-7< 3.\left(x-1\right)< 8\)
Vậy \(3.\left(x-1\right)\in\left\{-6;-5;-4;...;7\right\}\)
mà \(x\in Z\)nên \(3.\left(x-1\right)⋮3\)
Vậy \(3.\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)
hay \(x-1\in\left\{-2;-1;0;1;2\right\}\)
tới đây dễ rồi thì làm nốt nhé, để thời gian làm mấy câu sau!
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)
\(\Rightarrow1+\frac{1}{2.3}.2+\frac{1}{3.4}.2+...+\frac{1}{x\left(x+1\right)}.2=2\)
=> \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=2\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x\left(x+1\right)}=1\)
=> \(1-\frac{1}{x+1}=1\)
=> \(\frac{1}{x+1}=0\Rightarrow0\left(x+1\right)=1\Rightarrow x\in\varnothing\)
\(\frac{1}{1.2:2}+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{x.\left(x+1\right):2}=2\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=2\)
\(2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2\)
\(1-\frac{1}{x+1}=1\)
\(\frac{1}{x+1}=0\)
Vậy x vô nghiệm.