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29\(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\) + 39\(\dfrac{1}{3}\)\(\times\)\(\dfrac{3}{4}\) + \(\dfrac{5}{6}\)
= \(\dfrac{59}{2}\) \(\times\) \(\dfrac{2}{3}\) + \(\dfrac{118}{3}\) \(\times\) \(\dfrac{3}{4}\) + \(\dfrac{5}{6}\)
= \(\dfrac{59}{3}\) + \(\dfrac{59}{2}\) + \(\dfrac{5}{6}\)
= \(\dfrac{295}{6}\) + \(\dfrac{5}{6}\)
= 50
= 59/2 x 2/3+ 118/3 x 3/4 + 5/6
= 59/3+ 59/2+ 5/6
= 118/6+ 177/6+ 5/6
= 50
= 59/2 x 2/3+ 118/3 x 3/4 + 5/6
= 59/3+ 59/2+ 5/6
= 118/6+ 177/6+ 5/6
= 50
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
\(A=29\dfrac{1}{2}\cdot\dfrac{2}{3}+39\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)
\(=\dfrac{59}{2}\cdot\dfrac{2}{3}+\dfrac{118}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)
\(=\dfrac{59}{3}+\dfrac{118}{4}+\dfrac{5}{6}\)
\(=\dfrac{59}{3}+\dfrac{59}{2}+\dfrac{5}{6}\)
\(=59\cdot\left(\dfrac{1}{3}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\)
\(=\dfrac{5}{6}\cdot\left(59+1\right)=\dfrac{5}{6}\cdot60=50\)
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
b. \(\dfrac{13+x}{20}\)= \(\dfrac{15}{20}\)
=> 13+x=15
x=15-13
x=2
$#Shả$
`x xx1,2+x xx1,8=45`
`<=>x xx(1,2+1,8)=45`
`<=> x xx 3 =45`
`<=>x=45:3=15`
`(13+x)/20=3/4`
`<=>4xx(13+x)=20xx3`
`<=>4xx(13+x)=60`
`<=>13+x=60:4=15`
`<=>x=15-13=2`
a) \(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\)
\(\dfrac{2\times4}{3\times4}+\dfrac{3\times3}{4\times3}< x< \dfrac{\left(1\times3+1\right)\times5}{3\times5}+\dfrac{4\times3}{5\times3}\)
\(\dfrac{8}{12}+\dfrac{9}{12}< x< \dfrac{20}{15}+\dfrac{12}{15}\\ \dfrac{17}{12}< x< \dfrac{32}{15}\)
Ước tính: \(\dfrac{17}{12}=1,4\) và \(\dfrac{32}{15}=2,1\). Vậy số tự nhiên x = 2 sẽ thõa mãn 1,4 < x < 2,1
b)
\(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\\ \dfrac{5\times4}{6\times4}-\dfrac{1\times6}{4\times6}< x< \dfrac{\left(2\times3+1\right)\times5}{3\times5}-\dfrac{2\times3}{5\times3}\\ \dfrac{20}{24}-\dfrac{6}{24}< x< \dfrac{35}{15}-\dfrac{6}{15}\\ \dfrac{14}{24}< x< \dfrac{29}{15}\)
Ước tính \(\dfrac{14}{24}=0,5\) và \(\dfrac{29}{15}=1,9\)
Vậy với x là số tự nhiên x = 1 sẽ thõa mãn 0,5 < x < 1,9
\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{7}{12}\)
Muốn làm những bài toán có chứa hỗn số thì em cần nhớ:
Bước 1: Chuyển các hỗn số có trong phép tính thành phân số.
Bước 2: Thực hiện phép tính theo thứ tự thực hiện phép tính
a, \(\dfrac{8\dfrac{1}{2}}{15:\dfrac{5}{17}}\)
= \(\dfrac{\dfrac{17}{2}}{15\times\dfrac{17}{5}}\)
= \(\dfrac{\dfrac{17}{2}}{3\times17}\)
= \(\dfrac{17}{2}\) x \(\dfrac{1}{3\times17}\)
= \(\dfrac{1}{6}\)
`#3107`
`7/13 \div x = 14/39`
`=> x = 7/13 \div 14/39`
`=> x = 3/2`
Vậy, `x = 3/2`
________
`x \times 3/5 = 14/15`
`=> x = 14/15 \div 3/5`
`=> x = 14/9`
Vậy, `x=14/9.`
\(\dfrac{7}{13}:x=\dfrac{14}{39}\)
\(x=\dfrac{7}{13}:\dfrac{14}{39}\)
\(x=\dfrac{7}{13}\times\dfrac{39}{14}\)
\(x=\dfrac{3}{2}\)
___
\(x\times\dfrac{3}{5}=\dfrac{14}{15}\)
\(x=\dfrac{14}{15}:\dfrac{3}{5}\)
\(x=\dfrac{14}{15}\times\dfrac{5}{3}\)
\(x=\dfrac{14}{9}\)
\(\dfrac{2}{3}.\dfrac{39}{29}.\dfrac{2}{3}.\dfrac{29}{39}=\dfrac{2.39.2.29}{3.29.3.39}=\dfrac{2.2}{3.3}=\dfrac{4}{9}\)
\(\dfrac{2}{3}.\dfrac{39}{29}.\dfrac{2}{3}.\dfrac{29}{39}\)
=\(\dfrac{2}{3}.\dfrac{2}{3}.\left(\dfrac{39}{29}.\dfrac{29}{39}\right)\)
=\(\dfrac{4}{9}.1\)
=\(\dfrac{4}{9}\)