xét 5 trường hộp TH1 : x= 2007 TH2 : x= 2008 TH3 : x< 2007 TH4 : x > 2008 TH5 : 2007 < x < 2008 Vậy x \(\in\left\{2017,2018\right\}\)
Violympic toán 7
Các câu hỏi tương tự
Tìm x, y thỏa mãn :
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a, 2007\(\left|2x-y\right|\)2008 + 2008\(\left|y-4\right|\)2007 \(\le\)0
b, \(\left|5x+1\right|+\left|6y-8\right|\le0\)
Tìm x,y,z biết
\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
Tìm x,y,z biết
\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
Giúp mk vs!!!
Tim x, biet:
a, dfrac{x+1}{5}+dfrac{x+1}{6}dfrac{x+1}{7}+dfrac{x+1}{8}
b, dfrac{x-1}{2009}+dfrac{x-2}{2008}dfrac{x-3}{2007}+dfrac{x-4}{2006}
c, dfrac{3}{left(x+2right)left(x+5right)}+dfrac{5}{left(x+5right)left(x+10right)}+dfrac{7}{left(x+10right)left(x+17right)}dfrac{x}{left(x+2right)left(x+17right)}
d, dfrac{2}{left(x-1right)left(x-3right)}+dfrac{5}{left(x-3right)left(x-8right)}+dfrac{12}{left(x-8right)left(x-20right)}-dfrac{1}{x-20}dfrac{-3}{4}
Help me!!!!!!!!!!
Can gap lam. Ai lam duoc...
Đọc tiếp
Tim x, biet:
a, \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
b, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
c, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}\)\(=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
d, \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
Help me!!!!!!!!!!
Can gap lam. Ai lam duoc cau no thi lam nha. Cam on nhieu truoc!!!!!!!!!!!!
Tìm x,y,z biết
\(\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}=0\)
Tìm các giá trị của \(x,y\)thỏa mãn:
\(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)
tìm x biết :
\(\left|x-1\right|+2.\left|x-2\right|+3.\left|x-3\right|+4.\left|x-4\right|+5.\left|x-5\right|+20x=0\)
26 tháng 10 2021 lúc 12:36
Tìm x biết :
\(\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{4}\right|+....+\left|x-\dfrac{1}{10}\right|=2x\)
Tìm x, y biết :
\(\left|x+3\right|+\left|x-1\right|=\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\)