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Câu 1:
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\left(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}\right)\) - \(\left(\dfrac{x+1}{13}+\dfrac{x+1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)\)= 0
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
=> x = 0 - 1
=> x = -1
Câu 2:
Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-3.4+9+12}{n-4}\)
\(=\dfrac{3.\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để A có giá trị nguyên thì:
n - 4 \(\in\) Ư(21)
=> n - 4 \(\in\)
| n4 | 3 | -3 | 7 | -7 | -1 | 1 | -21 | 21 |
| n | 7 | 1 | 11 | -3 | 3 | 5 | -17 | 25 |
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.
\(\left\{{}\begin{matrix}1-\dfrac{2}{3}=\dfrac{1}{3}\\1-\dfrac{3}{4}=\dfrac{1}{4}\\1-\dfrac{4}{5}=\dfrac{1}{5}\\1-\dfrac{9}{10}=\dfrac{1}{10}\end{matrix}\right.\)
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
Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)
\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)
\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)
\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)
Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)
\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.
Vậy \(\dfrac{B}{A}\) là số nguyên.
Ta có: \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\Leftrightarrow a\left(b+n\right)< b\left(a+n\right)\)\(\Leftrightarrow ab+an< ab+bn\)\(\Leftrightarrow a< b\) (vì \(n>0\)).
Vậy \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\Leftrightarrow a< b.\)
Tương tự
\(\dfrac{a}{b}>\dfrac{a+n}{b+n}\Leftrightarrow a>b\) ;
\(\dfrac{a}{b}=\dfrac{a+n}{b+n}\Leftrightarrow a=b\).
1
a) Vì \(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Rightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\)
\(\Rightarrow ad< bc\)
2
b) Ta có : \(\dfrac{-1}{3}=\dfrac{-16}{48};\dfrac{-1}{4}=\dfrac{-12}{48}\)
Ta có dãy sau : \(\dfrac{-16}{48};\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48};\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa \(\dfrac{-1}{3}\) và \(\dfrac{-1}{4}\) là :\(\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}\)
1a ) Ta có : \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\)
\(\Leftrightarrow\) \(\dfrac{ad}{bd}\) < \(\dfrac{bc}{bd}\) \(\Rightarrow\) ad < bc
1b ) Như trên
2b) \(\dfrac{-1}{3}\) = \(\dfrac{-16}{48}\) ; \(\dfrac{-1}{4}\) = \(\dfrac{-12}{48}\)
\(\dfrac{-16}{48}\) < \(\dfrac{-15}{48}\) <\(\dfrac{-14}{48}\) < \(\dfrac{-13}{48}\) < \(\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa là.................
Cái này dễ mà em
a ) Để \(\dfrac{3}{n-1}\) là một số nguyên thì => 3 \(⋮\) (n - 1) hay n - 1 \(\in\) Ư (3) = { \(\pm\)1 , \(\pm\)3 }
=> n-1 = 1 => n= 2
n-1 = 3 => n= 4
n-1 = -1 => n= 0
n-1 = -3 => n= -2
Vậy n = 2 , n= -2 , n= 0 , n= 4
câu b ) tương tự nha em
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a\left(b+2017\right)}{b\left(b+2017\right)}\\\dfrac{a+2017}{b+2017}=\dfrac{b\left(a+2017\right)}{b\left(b+2017\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{ab+2017a}{b^2+2017b}\\\dfrac{a+2017}{b+2017}=\dfrac{ab+2017b}{b^2+2017b}\end{matrix}\right.\)
Ta cần so sánh:
\(ab+2017a\) với \(ab+2017b\)
Cần so sánh \(a\) với \(b\)
Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2017}{b+2017}\)
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2017}{b+2017}\)
Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2017}{b+2017}\)
Mấy câu sau dễ tương tự