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Bài 1.
Giải
a) Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-12+21}{n-4}=\dfrac{3\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để \(A\in Z\) thì \(\dfrac{21}{n-4}\in Z\)
\(\Rightarrow21⋮\left(n-4\right)\)
\(\Rightarrow\left(n-4\right)\inƯ\left(21\right)\)
\(\Rightarrow\left(n-4\right)\in\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Ta có bẳng sau:
| \(n-4\) | \(-21\) | \(-7\) | \(-3\) | \(-1\) | \(1\) | \(3\) | \(7\) | \(21\) |
| \(n\) | \(-17\) | \(-3\) | \(1\) | \(3\) | \(5\) | \(7\) | \(11\) | \(25\) |
Vậy \(n\in\left\{-17;-3;1;3;5;7;11;25\right\}\) thì \(A\in Z.\)
b) Ta có: \(B=\dfrac{6n+5}{2n-1}=\dfrac{6n-3+8}{2n-1}=\dfrac{3\left(2n-1\right)+8}{2n-1}=3+\dfrac{8}{2n-1}\)
Để \(B\in Z\) thì \(\dfrac{8}{2n-1}\in Z\)
\(\Rightarrow8⋮\left(2n-1\right)\)
\(\Rightarrow\left(2n-1\right)\inƯ\left(8\right)\)
\(\Rightarrow\left(2n-1\right)\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Ta có bảng sau:
| \(2n-1\) | \(-8\) | \(-4\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(4\) | \(8\) |
| \(2n\) | \(-7\) | \(-3\) | \(-1\) | \(0\) | \(2\) | \(3\) | \(5\) | \(9\) |
| \(n\) | \(\dfrac{-7}{2}\) | \(\dfrac{-3}{2}\) | \(\dfrac{-1}{2}\) | \(0\) | \(1\) | \(\dfrac{3}{2}\) | \(\dfrac{5}{2}\) | \(\dfrac{9}{2}\) |
Vậy \(n\in\left\{\dfrac{-7}{2};\dfrac{-3}{2};\dfrac{-1}{2};0;1;\dfrac{3}{2};\dfrac{5}{2};\dfrac{9}{2}\right\}\)
Bạn Nguyen Thi Huyen giải bài 1 rồi nên mình giải tiếp các bài kia nhé!
Bài 2:
\(\dfrac{x-18}{2000}+\dfrac{x-17}{2001}=\dfrac{x-16}{2002}+\dfrac{x-15}{2003}\)
\(\Leftrightarrow\left(\dfrac{x-18}{2000}-1\right)+\left(\dfrac{x-17}{2001}-1\right)=\left(\dfrac{x-16}{2002}-1\right)+\left(\dfrac{x-15}{2003}-1\right)\)
\(\Leftrightarrow\dfrac{x-2018}{2000}+\dfrac{x-2018}{2001}=\dfrac{x-2018}{2002}+\dfrac{x-2018}{2003}\)
\(\Leftrightarrow\dfrac{x-2018}{2000}+\dfrac{x-2018}{2001}-\dfrac{x-2018}{2002}-\dfrac{x-2018}{2003}=0\)
\(\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Dễ thấy \(\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}\) nên:
\(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\ne0\). Do đó:
\(x-2018=0\Leftrightarrow x=2018\)
Bài 3:
a) \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\Leftrightarrow\dfrac{20}{4x}+\dfrac{xy}{4x}=\dfrac{20+xy}{4x+4x}=\dfrac{20+xy}{8x}=\dfrac{1}{8}\)
Hoán vị ngoại tỉ ta có: \(\dfrac{20+xy}{8x}=\dfrac{1}{8}\Leftrightarrow\dfrac{8}{8x}=\dfrac{1}{x}=\dfrac{1}{8}\Leftrightarrow x=8\)
Thế x = 8 vào : \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\) .Ta có: \(\dfrac{5}{8}+\dfrac{y}{4}=\dfrac{1}{8}\Leftrightarrow\dfrac{y}{4}=\dfrac{1}{8}-\dfrac{5}{8}=\dfrac{-2}{4}\). Ta có: \(\dfrac{y}{4}=\dfrac{-2}{4}\Leftrightarrow y=-2\)
Vậy: \(\left[{}\begin{matrix}x=8\\y=-2\end{matrix}\right.\)
b) \(\dfrac{1}{x}-\dfrac{2}{y}=\dfrac{3}{1}\Rightarrow\dfrac{y}{x}-2=\dfrac{3}{1}\) (hoán vị ngoại tỉ)
\(\Leftrightarrow\dfrac{y}{x}=\dfrac{5}{1}\). Suy ra nghiệm x,y có dạng \(\left[{}\begin{matrix}x=1k\\y=5k\end{matrix}\right.\left(k\in Z\right)\). Bằng các phép thử lại ta dễ dàng suy ra x,y vô nghiệm.
a: =>x(1/2+1/4+1/2017)=x(1/3+1/5+1/2017)
=>x=0
b: =>1/-3=-7/21
e: a/b=2/7
nên a=2/7b
=>b=7/2a
b/c=14/15
=>b=14/15c
\(\Leftrightarrow\)7/2a=14/15c
=>a/c=4/15
\(\left\{{}\begin{matrix}1-\dfrac{2}{3}=\dfrac{1}{3}\\1-\dfrac{3}{4}=\dfrac{1}{4}\\1-\dfrac{4}{5}=\dfrac{1}{5}\\1-\dfrac{9}{10}=\dfrac{1}{10}\end{matrix}\right.\)
Vì:
\(\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}>...>\dfrac{1}{10}\)
nên:
\(\dfrac{2}{3}< \dfrac{3}{4}< \dfrac{4}{5}< ...< \dfrac{9}{10}\)
a)
Ta có:
\(\)\(\left\{{}\begin{matrix}\dfrac{3}{4}=\dfrac{2+1}{3+1}\\\dfrac{4}{5}=\dfrac{3+1}{4+1}\\\dfrac{5}{6}=\dfrac{4+1}{5+1}\\\dfrac{9}{10}=\dfrac{8+1}{9+1}\end{matrix}\right.\)
Suy ra quy luật:
Phân số tiếp theo chính là tử của p/s ban đầu +1/mẫu của p/s ban đầu +1
Vậy phân số sau phân số \(\dfrac{a}{b}\) là \(\dfrac{a+1}{b+1}\)
So sánh :
\(\dfrac{a}{b}\) và \(\dfrac{a+1}{b+1}\)
\(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\)
\(\dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\)
Vậy cần so sánh:
\(\dfrac{ab+a}{b^2+b}\) với \(\dfrac{ab+b}{b^2+b}\)
Cần so sánh:
\(ab+a\) và \(ab+b\)
Cần so sánh \(a\) với \(b\)
Nếu \(a>b\Rightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\)
Nếu \(a< b\Rightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)
Nếu \(a=b\) \(\Rightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}=1\)
Còn cách khác ngắn hơn nhưng lười làm lắm :v
a) 162n=2 => \(\dfrac{2^4}{2^n}=2\Rightarrow2^{4-n}=2\Rightarrow4-n=1\Rightarrow n=3\)
b,
\(\dfrac{\left(-3\right)^n}{81}=-27\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^4}=-27\Rightarrow\left(-3\right)^{n-4}=\left(-3\right)^3\Rightarrow n-4=3\Rightarrow n=7\)
c,\(8^n:2^n=4\Rightarrow4^n=4\Rightarrow n=1\)
=> (-3)n-4 = (-3)3
=> n - 4 = 3 => n = 7
c) 8n : 2n = 4
4n = 4.
a) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot2^5\)
\(\Rightarrow2^n\cdot4,5=288\)
\(\Rightarrow2^n=64\)
\(\Rightarrow n=6\)
b) \(2^m-2^n=1984\)
\(\Rightarrow2^n\cdot\left(2^{m-n}-1\right)=2^6\cdot31\)
\(\Rightarrow\left\{{}\begin{matrix}2^n=2^6\\2^{m-n}-1=31\end{matrix}\right.\)
\(\Rightarrow n=6\)
\(\Rightarrow2^{m-n}=32\Rightarrow m-n=5\Rightarrow m=11\)
@Ngô Tấn Đạt