KC
1
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Khách
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2 tháng 9 2021
Đặt a=4453, b=1997
Ta có: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+
\(=\dfrac{5a+6-4a-6}{ab}\)
\(=\dfrac{1}{b}\)
\(=\dfrac{1}{1997}\)
T
0
DC
1
16 tháng 9 2018
Đặt \(a=\frac{1}{4453};b=\frac{1}{1997}\)ta có :
\(5\frac{6}{4453}\cdot\frac{1}{1997}-\frac{2}{1997}\cdot2\frac{3}{4453}\)
\(=\left(5+6a\right)\cdot b-2b\left(2+3a\right)\)
\(=5b+6ab-4b-6ab\)
\(=b=\frac{1}{1997}\)
PC
1
KB
9 tháng 2 2019
Ta có :
\(\dfrac{1997^2-1996^2}{1997^2+1996^2}=\dfrac{1.\left(1997+1996\right)}{1997^2+1996^2}=\dfrac{3993}{1997^2+1996^2}\)
Lại có : \(\dfrac{1}{3993}=\dfrac{3993}{3993^2}\)
Do \(3993^2=\left(1997+1996\right)^2>1997^2+1996^2\)
\(\Rightarrow\dfrac{3993}{3993^2}< \dfrac{3993}{1997^2+1996^2}\)
\(\Rightarrow\dfrac{1}{3993}< \dfrac{1997^2-1996^2}{1997^2+1996^2}\)
14 tháng 12 2017
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)
14 tháng 12 2017
Vào trang cá nhân của t mà xem.T vừa làm r.Lười gõ lại lắm T^T
ST
14 tháng 12 2017
\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(=\dfrac{x+1}{1998}+\dfrac{x+2}{1997}-\dfrac{x+3}{1996}-\dfrac{x+4}{1995}=0\)
\(=\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1-\dfrac{x+3}{1996}-1-\dfrac{x+4}{1995}-1=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1998}-\left(\dfrac{x+3}{1996}+1\right)-\left(\dfrac{x+4}{1995}+1\right)=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(=\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
⇔\(x+1999=0\)
Vậy \(x=-1999\)
19 tháng 1 2019
\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)
\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)
Mấy câu khác tương tự :v
2 tháng 1 2023
b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
=>123-x=0
=>x=123
c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)
=>x-2019=0
=>x=2019
DH
1
31 tháng 10 2018
Sửa đề :
\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=-3\)
\(\Leftrightarrow\left(\dfrac{x+1}{2006}+1\right)+\left(\dfrac{x+10}{1997}+1\right)+\left(\dfrac{x+19}{1988}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+2007}{2006}+\dfrac{x+2007}{1997}+\dfrac{x+2007}{1988}=0\)
\(\Leftrightarrow\left(x+1007\right)\left(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\right)=0\)
Mà \(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\ne0\)
\(\Leftrightarrow x+2007=0\)
\(\Leftrightarrow x=-2007\)
Vậy..
Đặt 4453=a; 1997=b
\(A=\left(5+\dfrac{6}{a}\right)\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\left(2+\dfrac{3}{a}\right)\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)
\(=\dfrac{5a+6-4a-6}{ab}=\dfrac{a}{ab}=\dfrac{1}{b}=\dfrac{1}{1997}\)