<=> (x-1/2006 - 1)+(x-10/1997 - 1)+(x-19/1988 - 1) = 0
<=> x-2007/2006 + x-2007/1997 + x-2007/1988 = 0
<=> (x-2007).(1/2006+1/1997+1/1988) = 0
<=> x-2007=0 ( vì 1/2006+1/1997+1/1988 > 0 )
<=> x=2007
Vậy x=2007
k mk nha
tìm x
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
Tìm x bt
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
tìm x
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\)
tìm x:
x - \(\frac{1}{2006}\) + x - \(\frac{10}{1997}\) + x - \(\frac{19}{1988}\) = 3
tìm x biết
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
bài 1 :rút gọn A= (2+1).(22+1).(24+1).....(2256+1) + 1
bài 2 : a,cho x/a+y/b+z/c=0(1) và a/x+b/y+c/z=2 . tính A=x2/a2+y2/b2+z2/c2
b,rút gọn B=(ab/(a2+b2+c2))+(bc/(b2+c2+a2))+(ca/(c2+a2+b2))
bài 3: tìm x
((x-1)/2006)+((x-10)/1997)+((x-19)/1988)=3
Tính giá trị biểu thức sau
1, A=6+5^2+5^3+5^4 +....+5^1996+5^1997
2,B=10+9^2+9^3+9^4+....+9^2004+9^2005
3, C=x^20-2006x^19+x^2018-2006x^17+....+2006x^2-2006x+2006 với x=2005
giải pt
(x-13)/2006 + (x-22)/1997 + (x-31)/1998 =3
Giải Phương trình
\(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x^2\right)-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
