(6:\(\dfrac{3}{5}\)-1\(\dfrac{1}{6}\)x\(\dfrac{6}{7}\)) : (4\(\dfrac{1}{5}\)x\(\dfrac{10}{11}\)+5\(\dfrac{2}{11}\)) = ?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{6}x\dfrac{6}{7}}{\dfrac{4}{5}x\dfrac{10}{11}+5\dfrac{2}{12}}=\left(10-x\right):\left(\dfrac{8}{11}x+\dfrac{31}{6}\right)\)
\(\dfrac{10-\dfrac{7}{6}\cdot x\cdot\dfrac{6}{7}}{\dfrac{8}{11}\cdot x+\dfrac{31}{6}}=\dfrac{10-1\cdot x}{\dfrac{8}{11}\cdot x+\dfrac{31}{6}}=\dfrac{10-\dfrac{11}{11}\cdot x}{\dfrac{8}{11}\cdot x+\dfrac{31}{6}}=\dfrac{10-\dfrac{3}{11}\cdot x}{\dfrac{31}{6}}=>10-\dfrac{3}{11}\cdot x=\dfrac{31}{6} =>\dfrac{3}{11}\cdot x=10-\dfrac{31}{6}=\dfrac{29}{6}=>x=\dfrac{29}{6}:\dfrac{3}{11}=\dfrac{319}{18}\)
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
Bài này có cần phải tính nhanh ko vậy bn?
Nếu ko thì lấy máy tính mà tính cũng đc mà
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
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 \(B=3\)
\(3\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{10}{3}+\dfrac{5}{6}=\dfrac{20}{6}+\dfrac{5}{6}=\dfrac{25}{6}\)
\(4\dfrac{5}{7}-2\dfrac{11}{21}=\dfrac{33}{7}-\dfrac{53}{21}=\dfrac{99}{21}-\dfrac{53}{21}=\dfrac{46}{21}\)
\(3\dfrac{1}{4}x\dfrac{2}{5}=\dfrac{13}{3}x\dfrac{2}{5}=\dfrac{26}{15}\)
\(4\dfrac{7}{8}:\dfrac{1}{4}=\dfrac{39}{8}.4=\dfrac{39}{2}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
o