tìm x biết
\(\frac{26-x}{995}+\frac{22-x}{997}+\frac{18-x}{999}=\)\(\frac{12-x}{334}\)
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\(\frac{25-x}{995}+\frac{21-x}{997}+\frac{17-x}{999}=\frac{11-x}{334}\)
\(\Rightarrow\frac{25-x}{995}+2+\frac{21-x}{997}+\frac{17-x}{999}+2=\frac{11-x}{334}+6\)
\(\Rightarrow\frac{25-x}{995}+\frac{1990}{995}+\frac{21-x}{997}+\frac{1994}{997}+\frac{17-x}{999}+\frac{1998}{999}=\frac{11-x}{334}+\frac{2004}{334}\)
\(\Rightarrow\frac{2015-x}{995}+\frac{2015-x}{997}+\frac{2015-x}{999}=\frac{2015-x}{334}\)
\(\Rightarrow\frac{2015-x}{995}+\frac{2015-x}{997}+\frac{2015-x}{999}-\frac{2015-x}{334}=0\)
\(\Rightarrow\left(2015-x\right)\left(\frac{1}{995}+\frac{1}{997}+\frac{1}{999}+\frac{1}{334}\right)=0\)
\(\Rightarrow2015-x=0\left(\text{vì }\frac{1}{995}+\frac{1}{997}+\frac{1}{999}+\frac{1}{334}\ne0\right)\)
\(\Rightarrow x=2015\)
Câu 1:
a: =(1+2-3-4)+(5+6-7-8)+...+(2013+2014-2015-2016)
=(-4)+(-4)+...+(-4)
=-4x504=-2016
b: \(B=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{195}{196}=\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot13\cdot15}{2\cdot3\cdot...\cdot14\cdot2\cdot3\cdot...\cdot14}=\dfrac{15}{14\cdot2}=\dfrac{15}{28}\)
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
\(\frac{26-x}{995}\)+2+\(\frac{22-x}{997}\)+\(\frac{18-x}{999}\)+2=\(\frac{12-x}{334}\)+6
\(\frac{2016-x}{995.2}\)+\(\frac{2016-x}{997.2}\)+\(\frac{2016-x}{999.2}\)-\(\frac{2016-x}{334.6}\)=0
(2016-x)(\(\frac{1}{995.2}\)+\(\frac{1}{997.2}\)\(\frac{1}{999.2}\)-\(\frac{1}{334.6}\))=0
Dễ thấy giá trị biểu thức trong ngoặc thứ 2 khác 0
Nên 2016-x=0=>x=2016
Vậy giá trị của x là 2016