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\(\frac{7}{13}×\frac{5}{14}×\frac{39}{15}\)
\(=\frac{7×5×39}{13×14×15}\)
\(=\frac{7×5×13×3}{13×7×2×5×3}\)
\(=\frac{1}{2}\)
\(\frac{7}{13}×\frac{5}{14}×\frac{39}{15}\)
\(=\frac{7×5×39}{13×14×15}\)
\(=\frac{7×5×3×13}{13×2×7×3×5}\)
\(=\frac{1}{2}\)
`#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.`
\(\dfrac{7}{13}:x=\dfrac{14}{39}\)
\(x=\dfrac{7}{13}:\dfrac{14}{39}\)
\(x=\dfrac{7}{13}\times\dfrac{39}{14}\)
\(x=\dfrac{3}{2}\)
___
\(x\times\dfrac{3}{5}=\dfrac{14}{15}\)
\(x=\dfrac{14}{15}:\dfrac{3}{5}\)
\(x=\dfrac{14}{15}\times\dfrac{5}{3}\)
\(x=\dfrac{14}{9}\)
\(\frac{7}{13}\cdot\frac{5}{14}\cdot\frac{39}{15}\)
Theo đề bài ta có :
\(\Leftrightarrow\frac{7.5.39}{13.14.15}\Leftrightarrow\frac{1.1.3}{1.2.3}\Leftrightarrow\frac{3}{6}\Leftrightarrow\frac{1}{2}\)
:) :) :) :)
a)<=>2/7:(13/24+5/24) b)<=>15/24-9/24-4/24 c)=7.5.39/13.14.15
<=>2/7:3/4 <=>2/24=1/12 =3/2.3
<=>2/7x4/3 =1/2
<=>8/21
a. \(\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
= \(\frac{2}{7}:\left(\frac{13}{24}+\frac{5}{24}\right)\)
= \(\frac{2}{7}:\frac{3}{4}\)
= \(\frac{8}{21}\)
b. \(\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
= \(\frac{15}{24}-\frac{9}{24}-\frac{4}{24}\)
= \(\frac{2}{24}=\frac{1}{12}\)
c. \(\frac{7}{13}.\frac{5}{14}.\frac{39}{15}\)
= \(\frac{7.5.3.13}{13.2.7.3.5}\)
= \(\frac{1}{2}\)
bạn ơi bạn bị sai mất 2 câu cuối rồi đó làm ơn tính lại giùm mình nha
1.3.77−1+3.7.99−3+7.9.1313−7+9.13.1515−9+\frac{19-13}{13.15.19}+13.15.1919−13
=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31−3.71+3.71−7.91+7.91−9.131+9.131−13.151+13.151−15.191
=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31−15.191=28595−2851=28594
b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61.(1.3.76+3.7.96+7.9.136+9.13.156+13.15.196)
làm giống như trên
c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81.(1.2.31+2.3.41+3.4.51+...+48.49.501)
=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161.(1.2.32+2.3.42+3.4.52+...+48.49.502)
=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161.(1.2.33−1+2.3.44−2+3.4.55−3+...+48.49.5050−48)
=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161.(1.21−2.31+2.31−3.41+3.41−4.51+...+48.491−49.501)
=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161.(21−24501)=161.(24501225−24501)=4900153
d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75.(1.5.87+5.8.127+8.12.157+...+33.36.407)
=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75.(1.5.88−1+5.8.1212−5+8.12.1515−8+...+33.36.4040−33)
=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75.(1.51−5.81+5.81−8.121+8.121−12.151+...+33.361−36.401)
=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75.(51−14401)=75.(1440288−14401)=28841
P/S: . là nhân nha
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
\(\dfrac{7}{13}\times\dfrac{5}{14}\times\dfrac{39}{15}=\dfrac{7\times5\times3\times13}{13\times7\times2\times5\times3}=\dfrac{1}{2}\)
\(\dfrac{7}{13}\cdot\dfrac{5}{14}\cdot\dfrac{39}{15}=\dfrac{1}{2}\)