Tìm số tự nhiên x thỏa mãn:
\(\frac{1}{2016-x}+\frac{1}{2017-x}+\frac{1}{2018-x}=\frac{13}{12}\)
cho x,y,z thỏa mãn \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)
tìm B=\(\left(x^{2016}+y^{2016}\right)\left(y^{2017}+z^{2017}\right)\left(z^{2018}+x^{2018}\right)\)
Tìm số tự nhiên x thỏa mãn: \(\frac{1}{2014-x}\)+ \(\frac{1}{2015-x}\)+ \(\frac{1}{2016-x}\)= \(\frac{13}{12}\)
Tìm x thỏa mãn \(\frac{x}{2016}+\frac{x+1}{2017}=\frac{x+2}{2018}\)\(+\frac{x+3}{2019}\)
Dễ thấy bài này kết quả bằng 2016 rồi nhưng mình cần cách giải
Ai nhanh và đúng sẽ nhận tích nhe !
Tìm x biết:
a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Leftrightarrow\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}>0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Leftrightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Leftrightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
a) \(\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
=>\(x+2=0\)
=>\(x=-2\)
nếu có sai thì mong bn thông cảm nha
BÀI 1: tìm x biết : \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
BÀI 2: tìm số tự nhiên x thỏa mãn: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{16}{34}\)
BÀI 3: Cho x;y thỏa mãn : \(\left(x-2014\right)^{2010}+\left(y-2010\right)^{2014}\le0\)
bài 1
[(x+2)/1010]+ [(x+2)/1111]= [(x+2)/1212]+[(x+2)/1313]
=>[(x+2)/1010]+[(x+2)/1111] - [(x+2)/1212]-[(x+2)/1313] = 0
=>(x+2).[(1/1010)+(1/1111)-(1/1212)-(1/1313)=0
Vì [(1/1010)+(1/1111)-(1/1212)-(1/1313)] khác 0
=>x+2=0
=>x=-2
Bài 1 : -2
Bài 2 : 15
Bải 3 : x =2014 ; y = 2010
Cho các số thực x,y,z thỏa mãn\(x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=6\)
Tính A= x2016+y2017+z2018
Tìm x
\(\frac{x-2017}{2015.2016}+\frac{x-2018}{2016.2017}+\frac{x-2019}{2017.2018}+\frac{x-2020}{2018.1019}=\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{1018}\)
Tìm x , y thỏa mãn :
a) \(\frac{1}{2}\times(\frac{3}{4}x-\frac{1}{2})^{2018}+\frac{2017}{2018}\times/\frac{4}{5}y+\frac{6}{25}/\le0\)0
b) \(2017\times/2x-y/+2018\times(y-4)^{2017}\le0\)
Tìm x\(\inℝ\)biết:
\(\frac{x+1}{2018}=\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{3x+12}{2015}\)