B = 1/4 x( 4/ 1x3x5 +4 /3x5x7 +...+ 4/47x49x51)

= 1/4 x( 1/ 1x3 - 1/3x5 +1/3x5 -1/5x7 +...+ 1/47x49 -1/49x51)

= 1/4 x( 1/1x3 -1/49x51)

= 1/4 x( 1/3 - 1/2499)

= 1/4 x 832/2499

= 208 /2499

Vậy B= 208 /2499

C= 1/ 9x11 -1/ 11x13 +1/ 11x13 -1/ 13x15 +... + 1/ 59x61 -1/61x63

= 1/ 9x11 -1/ 61x63

= 1/99 -1/3843

= 416 /42273

Vậy C= 416 /42273

Mình ko hieu ý thu 2

a. 49.51- 49.4 -51.49 -51.4

= -49.4 - 51.4

=4.(-49-51)

=4.-100

=-400

b.71.64 + 32.(-7) -13.32

= 71.64 + 32.(-7-13)

=71.64 + 32.-20 

=71.32.2 +32.-20

=32.(71x2-20)

=32. 122

= 3904

c. (11-13)+(15-17)+..+(59-61)

=-2 x 13

=-26

\(A=\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{59\cdot61}=2\cdot\left(\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{59\cdot61}\right)\)

\(=2\cdot\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)=2\cdot\left[\left(\frac{1}{7}-\frac{1}{61}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{59}-\frac{1}{59}\right)\right]\)\(=2\cdot\left[\left(\frac{61}{427}-\frac{7}{427}\right)+0+...+0\right]=2\cdot\frac{54}{427}=\frac{108}{427}\)

Cách làm thì đúng rồi nhưng mk k chắc tính đúng, nhẩm thôi

xl bạn nhé nhưng bạn hiểu nhầm r ( chắc do mình ) nó là \(\frac{4\cdot9}{7}\)nhé bạn =))

a)

\( 49.(51 - 4) - 51.(49 + 4)\)

\(=49.51-49.4-51.49-51.4\)

\(=49.(51-4-51)-51.4\)

\(=49.(-4)-51.4\)

\(=4.(-49-51)\)

\(=4.(-100)\)

\(=-400\)

b)

\( 71.64 + 32.(-7) - 13.32\)

\(=71.32.2+32.(-7)-13.32\)

\(=142.32+32.(-7)-13.32\)

\(=32.(142-7-13)\)

\(=32.122\)

\(=3904\)

c)

\( 11 + (-13) + 15 + (-17) + ... + 59 + (-61)\)

\(=(-2)+(-2)+...+(-2)\)

\(=(-2).13\)

\(=-26\)

\(\frac{4}{9.11}+\frac{4}{11.13}+\frac{4}{13.15}+...+\frac{4}{97.99}\)

\(=2\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{9}-\frac{1}{99}\right)=\frac{20}{99}\)