Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(1+\frac{1}{2}\right)-1+\frac{1}{6}+\left(\frac{1}{2}+\frac{1}{12}\right)-\frac{1}{2}+\frac{1}{20}+\left(\frac{1}{3}+\frac{1}{30}\right)-\frac{1}{3}+\frac{1}{42}+\left(\frac{1}{4}+\frac{1}{56}\right)-\frac{1}{4}+\frac{1}{72}\)
=\(=\left(1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)
\(=0+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{8}-\frac{1}{8}\right)\)\(=\left(\frac{9}{9}-\frac{1}{9}\right)+0+...+0=\frac{8}{9}\)
\(M=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{17}{72}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)
\(=\dfrac{108}{72}-\dfrac{60}{72}+\dfrac{42}{72}-\dfrac{17}{72}+\dfrac{-27}{60}+\dfrac{22}{60}+\dfrac{-52}{168}+\dfrac{45}{168}\)
\(=\dfrac{73}{72}-\dfrac{1}{12}-\dfrac{1}{24}=\dfrac{73}{72}-\dfrac{6}{72}-\dfrac{3}{72}=\dfrac{64}{72}=\dfrac{8}{9}\)
ta có : ( -5/28 +7/4 + 8/35 ) : (- 69/20)
= ( -25/140 + 245/140 + 32/140 ) x (-20/69)
= (252/140) x (-20/69)
= (9/5) x (-20/69)
= (- 12/23)
tính nhanh:
2 x 3/7 + (2/9 - 10/7) - 5/3 x 9
= 6/7 + 2/9 - 10/7 - 5/3 x 9 = 6/7 + 2/9 - 10/7 - 15
= (6/7 - 10/7 ) + (2/9 - 135/9) = ( - 4/7 ) + (-133/9 )
= (- 36/63) + (-931/63)
= (- 967/63)
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{12}{7}=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}=\frac{-5}{7}.1+\frac{12}{7}=\frac{-5}{7}+\frac{12}{7}=-1\)
B=522+313−12413−211+32
Đặt \(S=\frac{5}{22}+\frac{3}{13}-\frac{1}{2}\)
=> S= \(\frac{-6}{143}\)
Đặt \(H=\frac{4}{13}-\frac{2}{11}+\frac{3}{2}\)
=> H = \(\frac{465}{286}\)
Khi đó B = \(\frac{\frac{-6}{143}}{\frac{465}{286}}\)=\(\frac{-4}{155}\)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
giúp giùm mình đi