X x 1999 - x = 1999 x 1997 + 1999

X x (1999 - 1) = 1999 x (1997 + 1)

X x 1998 = 1999 x 1998

Vậy x = 1999

2x1999-5=1999x1999+1999

x=1999

nha

X * 1999 - x = 1999 x 1997 + 1999

X * 1998      = 1999 x 1998

=> x = 1999

1999x-(-x)=1999.1997+1999

<=>2000x=3994002

<=>x=1997,001

X.1999.X=1999.1997+1999

X.1999.X=1999.1997+1999.1

X.X.1999=1999.(1997+1)

X.X.1999=1999.1998

X.X=1999.1998:1999=1998

 X × 1999 × X = 1999 × 1997 + 1999

X^2 x 1999 = 1999 × 1997 + 1999 x 1

X^2 x 1999 = 1999 x ( 1997 + 1 )

X^2 x 1999 = 1999 x 1998

X^2 = 1999 x 1998 : 1999

X^2 = 1999 : 1999 x 1998

X^2 = 1 x 1998

X^2 = 1998

a , x = 501

b , x = 1999

a) 5 * x - 1952 = 2500 - 1947

    5 * x -1952 =553

    5 * x             = 553 + 1952

    5 * x             =2505

    x                   = 501

b)x * 1999 - x = 1999*1997 +1999

 x * ( 1999 -1) = 1999*(1997 +1)

 x * 1998          =1999 * 1998

x                        =1999

X . 1999 - x = 1999 x 1997 + 1999

X . 1999 - X x 1 = 1999 x 1997 + 1999 x 1

X . 1997 = 1999 x 1998

X . 1997 = 3994002

X = 3994002 : 1997

X = 1997001

x . 1999 - x = 1999 . 1997 + 1999

=> x.(1999 - 1) = 1999.(1997 + 1)

=> x.1998 = 1999.1998

=> x = 1999

vậy x = 1999

Dấu . là nhân nha

X x 1999 - x = 1999 x 1997 + 1999

X x 1999 - X x 1 = 1999 x 1997 + 1999 x 1

X x (1998 - 1) = 1999 x (1997 + 1)

X x 1997 = 1999 x 1998

X x 1997 = 3994002

X = 3994002 : 1997

X = 1997001

3992006 phan 7980012

1000 nha

cảm ơn bạn nha

Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)

\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)

\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)

\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)

\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)

\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)

Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)

=> x - 2000 = 0 

=> x = 2000