Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Để \(\frac{6}{2a+1}\inℤ\)thì \(6⋮2a+1\)
\(\Rightarrow2a+1\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Vì \(a\inℤ\)\(\Rightarrow2a+1\)là số lẻ
\(\Rightarrow\)\(2a+1\)là ước lẻ của 6
\(\Rightarrow2a+1\in\left\{-3;-1;1;3\right\}\)
\(\Rightarrow2a\in\left\{-4;-2;0;2\right\}\)
\(\Rightarrow a\in\left\{-2;-1;0;1\right\}\)
Vậy \(a\in\left\{-2;-1;0;1\right\}\)
b) Để \(\frac{4a-3}{5a-1}\inℤ\)thì \(4a-3⋮5a-1\)\(\Rightarrow5.\left(4a-3\right)⋮5a-1\)
Ta có: \(5\left(4a-3\right)=20a-15=20a-4-11=4\left(5a-1\right)-11\)
Vì \(4.\left(5a-1\right)⋮5a-1\)\(\Rightarrow\)Để \(4a-3⋮5a-1\)thì \(11⋮5a-1\)
\(\Rightarrow5a-1\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Leftrightarrow5a\in\left\{-10;0;2;12\right\}\)\(\Leftrightarrow a\in\left\{-2;0;\frac{2}{5};\frac{12}{5}\right\}\)
mà \(a\inℤ\)\(\Rightarrow a\in\left\{-2;0\right\}\)
Vậy \(a\in\left\{-2;0\right\}\)
c) \(\frac{a^2+3}{a-1}=\frac{a^2-1+4}{a-1}=\frac{\left(a-1\right)\left(a+1\right)+4}{a-1}=\left(a+1\right)+\frac{4}{a-1}\)
Vì \(a\inℤ\)\(\Rightarrow a+1\inℤ\)
\(\Rightarrow\)Để \(\frac{a^2+3}{a-1}\inℤ\)thì \(\frac{4}{a-1}\inℤ\)
\(\Rightarrow4⋮a-1\)\(\Rightarrow a-1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow a\in\left\{-3;-1;0;2;3;5\right\}\)
Vậy \(a\in\left\{-3;-1;0;2;3;5\right\}\)
\(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)
\(\frac{2}{3}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{2}{3}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}:\frac{2}{3}\)
\(x=-\frac{3}{8}\)
\(\frac{2}{3}x\)\(=\)\(\frac{1}{4}\)\(-\)\(\frac{1}{2}\)
\(\frac{2}{3}x\)\(=\)\(\frac{-1}{4}\)
\(x\)\(=\)\(\frac{-1}{4}\)\(:\)\(\frac{2}{3}\)
\(x\)\(=\)\(\frac{-3}{8}\)
Ta có A = \(133\left(\frac{1}{1.1996}+\frac{1}{2.1997}+...+\frac{1}{17.2002}\right)\)
=> 1995A = \(133\left(\frac{1995}{1.1996}+\frac{1995}{2.1997}+...+\frac{1995}{17.2002}\right)\)
=> 1995A = \(133\left(1-\frac{1}{1996}+\frac{1}{2}-\frac{1}{1997}+...+\frac{1}{17}-\frac{1}{2002}\right)\)
=> 1995A = \(133\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2002}\right)\right]\)
=> A = \(\frac{1}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2002}\right)\right]\)(1)
Lại có B = \(\frac{17}{15}\left(\frac{1}{1.18}+\frac{1}{2.19}+...+\frac{1}{1995.2012}\right)\)
=> 17B = \(\frac{17}{15}\left(\frac{17}{1.18}+\frac{17}{2.19}+...+\frac{17}{1995.2012}\right)\)
=> 17B = \(\frac{17}{15}\left(1-\frac{1}{18}+\frac{1}{2}-\frac{1}{19}+...+\frac{1}{1995}-\frac{1}{2012}\right)\)
=> 17B = \(\frac{17}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{1995}\right)-\left(\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2012}\right)\right]\)
=> 17B = \(\frac{17}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2012}\right)\right]\)
=> B = \(\frac{1}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2012}\right)\right]\)(2)
Từ (1) và (2) => A = B
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
Khử mẫu : \(12x-12+6x-6=4x+3x-7\)
\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)
\(\Leftrightarrow18x-18=7x-7\)
\(\Leftrightarrow18x+7x=18+7\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)
Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)
\(\Leftrightarrow x=1\)
a) \(\frac{1-x}{x+4}=\frac{5-4-x}{x+4}=\frac{5}{x+4}-1\inℤ\Leftrightarrow\frac{5}{x+4}\inℤ\)
mà \(x\inℤ\Rightarrow x+4\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)
\(\Leftrightarrow x\in\left\{-9,-5,-3,1\right\}\)
b) \(\frac{11-2x}{x-5}=\frac{1+10-2x}{x-5}=\frac{1}{x-5}-2\inℤ\Leftrightarrow\frac{1}{x-5}\inℤ\)
mà \(x\inℤ\Rightarrow x-5\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{4,6\right\}\)
c) \(\frac{x+1}{2x+1}\inℤ\Rightarrow\frac{2\left(x+1\right)}{2x+1}=\frac{2x+1+1}{2x+1}=1+\frac{1}{2x+1}\inℤ\Leftrightarrow\frac{1}{2x+1}\inℤ\)
mà \(x\inℤ\Rightarrow2x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-1,0\right\}\).
Thử lại đều thỏa mãn.
Ta có 4A=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
Trừ 4A cho A ta được
3A = \(1-\frac{1}{2^{100}}\)=> 3A <1 => A<1/3 (đpcm)
Chúc bạn học tốt
Ta có :\(A=\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(2A=\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{100}}\)
Lại có :
\(\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)
Vì \(\frac{1}{2^{100}}< \frac{1}{6}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2^{100}}>\frac{1}{2}-\frac{1}{6}\)
\(\Rightarrow A>\frac{1}{3}\)
Vậy \(A>\frac{1}{3}\)(ĐPCM)
Ta có :\(\frac{1}{a}+\frac{1}{b}=\frac{4}{ab}-1\)
\(\Rightarrow\frac{a+b}{ab}=\frac{4}{ab}-1\)
=> \(\frac{a+b-4}{ab}=-1\)
=> a + b - 4 = -ab
=> a + b - 4 + ab = 0
=> a(b + 1) + b + 1 - 5 = 0
=> (a + 1)(b + 1) = 5
Vì \(a;b\inℤ\Rightarrow\hept{\begin{cases}a+1\inℤ\\b+1\inℤ\end{cases}}\)
Khi đó 5 = 1.5 = (-1).(-5)
Lập bảng xét các trường hợp
Vậy các cặp (a;b) nguyên thỏa mãn là (-6 ; -2) ; (-2 ; -6)
\(\frac{1}{a}+\frac{1}{b}=\frac{4}{ab}-1\)( ĐKXĐ : \(a,b\ne0\)) ( Bạn Xyz nhớ bổ sung thêm ĐKXĐ ạ )
\(\Leftrightarrow\frac{b}{ab}+\frac{a}{ab}=\frac{4}{ab}-\frac{ab}{ab}\)
\(\Leftrightarrow\frac{b}{ab}+\frac{a}{ab}-\frac{4}{ab}+\frac{ab}{ab}=0\)
\(\Leftrightarrow\frac{b+a-4+ab}{ab}=0\)
\(\Leftrightarrow b+a-4+ab=0\)
\(\Leftrightarrow b+a-5+1+ab=0\)
\(\Leftrightarrow a\left(b+1\right)+1\left(b+1\right)=5\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)=5\)
Ta có bảng sau :
Theo ĐKXĐ => Các cặp ( x; y ) thỏa mãn là : ( -2 ; -6 ) ; ( -6 ; -2 )