7/11+6/5
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`a, 2/7 +(3/7 + 5/7)= 2/7 + 3/7 + 5/7=(2/7 +5/7)+3/7= 1+3/7=7/7+3/7=10/7`
`b, (6/11+5/11)+4/11= 11/11+4/11=15/11`
`c, (5/6 + 7/6)+2/6= 12/6 + 2/6= 14/6= 7/3`
`@ yl`
\(a\) \(\dfrac{2}{7}+\left(\dfrac{3}{7}+\dfrac{5}{7}\right)=\dfrac{2}{7}+\dfrac{8}{7}=\dfrac{10}{7}\)
\(b\) \(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{4}{11}=\dfrac{11}{11}+\dfrac{4}{11}=\dfrac{15}{11}\)
\(c\) \(\left(\dfrac{5}{6}+\dfrac{7}{6}\right)+\dfrac{2}{6}=\dfrac{12}{6}+\dfrac{2}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)
\(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
\(\dfrac{5}{11}.\left(-\dfrac{3}{7}\right)+\dfrac{5}{11}.\left(-\dfrac{5}{7}\right)+\left(-\dfrac{8}{7}\right).\dfrac{6}{11}\)
\(=\dfrac{5}{11}.\left[-\dfrac{3}{7}+\left(-\dfrac{5}{7}\right)\right]+\left(-\dfrac{8}{7}\right).\dfrac{6}{11}\)
\(=\dfrac{5}{11}.\left(-\dfrac{8}{7}\right)+\left(-\dfrac{8}{7}\right).\dfrac{6}{11}=\left(\dfrac{5}{11}+\dfrac{6}{11}\right).\left(-\dfrac{8}{7}\right)\)
\(=1.\left(-\dfrac{8}{7}\right)=-\dfrac{8}{7}\)
a, \(\dfrac{5}{9}.\dfrac{10}{11}+\dfrac{5}{9}.\dfrac{14}{11}-\dfrac{5}{9}.\dfrac{15}{11}=\dfrac{5}{9}.\left(\dfrac{10}{11}+\dfrac{14}{11}-\dfrac{15}{11}\right)=\dfrac{5}{9}.\dfrac{9}{11}=\dfrac{5}{11}\)
b, \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{3}{13}.\dfrac{6}{7}\)\(=\dfrac{6}{7}.\left(\dfrac{8}{13}-\dfrac{3}{13}\right)+\dfrac{6}{13}.\dfrac{9}{7}=\dfrac{6}{7}.\dfrac{5}{13}+\dfrac{54}{91}=\dfrac{30}{91}+\dfrac{54}{91}=\dfrac{84}{91}=\dfrac{12}{13}\)
Bạn ơi, gõ latex cho dễ nhìn nhé!
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
\(\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{11}\cdot\dfrac{1}{7}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{7}\cdot\dfrac{1}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{7}\cdot\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}\cdot\dfrac{-7}{11}=\dfrac{-5}{11}\)
a ,b
cách 1 : cộng 2 số trong ngặc rồi tính bình thường
cách 2 : nhân phân phối ra rồi tính nhân chia trước cộng trừ sau
c,d
cách 1 : tính bình thường, nhân chia trc cộng trừ sau
cách 2 : đặt nhân tử ( thừa số ) chung
Cách 1 : \((\frac{6}{11}+\frac{5}{11})\cdot\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}\)
Cách 2 : \(\frac{6}{11}\cdot\frac{3}{7}+\frac{5}{11}\cdot\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)
Cách 1 dễ rồi nha
Cách 2 : \(\frac{6}{7}:\frac{2}{5}-\frac{4}{7}:\frac{2}{5}=\frac{15}{7}-\frac{10}{7}=\frac{5}{7}\)
Cách 1 tự làm
Cách 2 : \(\frac{3}{5}\cdot(\frac{7}{9}-\frac{2}{9})=\frac{3}{5}\cdot\frac{5}{9}=\frac{1}{3}\)
Cách 1 : ...
Cách 2 : \((\frac{8}{15}+\frac{7}{15}):\frac{2}{11}=1:\frac{2}{11}=\frac{2}{11\cdot1}=\frac{2}{11}\)
b: \(27D=3^{14}+3^{17}+...+3^{2024}\)
\(\Leftrightarrow26D=3^{2024}-3^{11}\)
hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)
c: \(25E=-5^4-5^6-...-5^{1002}\)
\(\Leftrightarrow24E=-5^{1002}+5^2\)
hay \(E=\dfrac{-5^{1002}+5^2}{24}\)
`7/11+6/5=35/55+66/55=101/55`
`7/11+6/5`
`=35/55+66/55`
`=101/55`