Question

TÌM  SỐ TỰ NHIÊN X : 

1/3 + 1/6 + 1/10 + ...  1/ x × ( x + 1 ) : 2 = 2001/ 2003

TÌM X : 

a) ( x + 1 ) \(^{ }^2\)= 4/3 × 75/9

b ) ( 4,5x  - 2x ) × ( -11/ 7 ) = 11/14

Answer 1

a) (x + 1)² = 4/3. 75/9

=> (x + 1)² = 100/9

=> (x + 1)² = (10/3)²

=> \(\orbr{\begin{cases}x+1=\frac{10}{3}\\x+1=-\frac{10}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{13}{3}\end{cases}}\)

Answer 2

b) (4,5x - 2x) . (-11/7) = 11/14

=> 2,5x = 11/14 : (-11/7)

=> 2,5x = -1/2

=> x = -1/2 : 2,5

=> x = -0,2

Answer 3

#)Giải :

Tìm số tự nhiên X :

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}:2=\frac{2001}{2003}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(\Rightarrow x=2003-1\)

\(\Rightarrow x=2002\)

                #~Will~be~Pens~#

Answer 4

 1/3 + 1/6 + 1/10 + ... + 1/x . ( x + 1 ) : 2 = 2001/2003

= 2/2.3 + 2/3.4 + 2/4.5 + ... + 2/x ( x + 1 )

= 2[ 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/ ( x + 1 ) ]

= 2[ 1/2 - 1/x + 1] = ( x - 1 ) / ( x + 1 ) = 2001/2003

=> x = 2002