\(\dfrac{423}{2813}\)-\(\dfrac{5}{6}\)=
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\(\dfrac{423}{2813}\)-\(\dfrac{5}{6}\)=
\(\dfrac{423}{2813}-\dfrac{5}{6}=\dfrac{2538}{16878}+\dfrac{14065}{16878}=\dfrac{16603}{16878}\)
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Bài 1 - Đặt tính rồi tính15 423 : 9 306 427 : 23706 999 : 201 7 406 x 236 200 x 2069 731 x 27Bài 2 - Tính giá trị của biểu thứca, dfrac{2}{5} + dfrac{2}{7} x dfrac{2}{3}b, dfrac{2}{6} : dfrac{7}{5} + dfrac{2}{6}c, ( dfrac{2}{3} - dfrac{1}{6} ) x dfrac{6}{7}Giúp tớ nhé!!!!
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Bài 1 - Đặt tính rồi tính
15 423 : 9
306 427 : 23
706 999 : 201
7 406 x 2
36 200 x 206
9 731 x 27
Bài 2 - Tính giá trị của biểu thức
\(a,\) \(\dfrac{2}{5}\) + \(\dfrac{2}{7}\) x \(\dfrac{2}{3}\)
\(b,\) \(\dfrac{2}{6}\) : \(\dfrac{7}{5}\) + \(\dfrac{2}{6}\)
\(c,\) \((\) \(\dfrac{2}{3}\) - \(\dfrac{1}{6}\) \()\) x \(\dfrac{6}{7}\)
Giúp tớ nhé!!!!
em đăng bên box toán để mấy bạn kia trả lời nhé =)))))
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Giải phương trình sau
\(\dfrac{x-2837}{824}+\dfrac{x-2813}{800}+\dfrac{x-1979}{34}+\dfrac{x-1991}{22}=0\)
Giải:
\(\dfrac{x-2837}{824}+\dfrac{x-2813}{800}+\dfrac{x-1979}{34}+\dfrac{x-1991}{22}=0\)
\(\Leftrightarrow\dfrac{x-2837}{824}+\dfrac{x-2813}{800}+\dfrac{x-1979}{34}+\dfrac{x-1991}{22}+0=0\)
\(\Leftrightarrow\dfrac{x-2837}{824}+\dfrac{x-2813}{800}+\dfrac{x-1979}{34}+\dfrac{x-1991}{22}+2-2=0\)
\(\Leftrightarrow\dfrac{x-2837}{824}+1+\dfrac{x-2813}{800}+1+\dfrac{x-1979}{34}-1+\dfrac{x-1991}{22}-1=0\)
\(\Leftrightarrow\dfrac{x-2837+824}{824}+\dfrac{x-2813+800}{800}+\dfrac{x-1979-34}{34}+\dfrac{x-1991-22}{22}=0\)
\(\Leftrightarrow\dfrac{x-2013}{824}+\dfrac{x-2013}{800}+\dfrac{x-2013}{34}+\dfrac{x-2013}{22}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{824}+\dfrac{1}{800}+\dfrac{1}{34}+\dfrac{1}{22}\right)=0\)
Vì \(\dfrac{1}{824}+\dfrac{1}{800}+\dfrac{1}{34}+\dfrac{1}{22}\ne0\)
Nên \(x-2013=0\)
\(\Leftrightarrow x=2013\)
Vậy \(x=2013\)
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\(\dfrac{1}{423}\):45,978
Tính nhanh:
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\)\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\)
giúp mình với
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
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 \(B=3\)
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Tính A dfrac{1}{2}- dfrac{2}{3}+dfrac{3}{4}-dfrac{4}{5}+dfrac{5}{6}-dfrac{6}{7}-dfrac{6}{5}+dfrac{4}{5}-dfrac{3}{4}+dfrac{2}{3}-dfrac{1}{2}
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Tính A= \(\dfrac{1}{2}\)- \(\dfrac{2}{3}\)+\(\dfrac{3}{4}\)-\(\dfrac{4}{5}\)+\(\dfrac{5}{6}\)-\(\dfrac{6}{7}\)-\(\dfrac{6}{5}\)+\(\dfrac{4}{5}\)-\(\dfrac{3}{4}\)+\(\dfrac{2}{3}\)-\(\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}\right)\)
\(=0+0+0+0-\dfrac{257}{210}\)
\(=\dfrac{257}{210}\)
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Có ai biết câu này không, làm giúp mình với
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SGK Cánh diều - Tập 2 - Trang 49
28 tháng 10 2023 lúc 0:32
Tính rồi rút gọn (theo mẫu):Mẫu: dfrac{5}{6}+dfrac{4}{6}dfrac{5+4}{6}dfrac{9}{6}dfrac{3}{2}a) dfrac{1}{8}+dfrac{5}{8} b) dfrac{1}{15}+dfrac{4}{15} c) dfrac{5}{9}+dfrac{7}{9} d) dfrac{23}{100}+dfrac{27}{100}
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Tính rồi rút gọn (theo mẫu):
| Mẫu: \(\dfrac{5}{6}+\dfrac{4}{6}=\dfrac{5+4}{6}=\dfrac{9}{6}=\dfrac{3}{2}\) |
a) \(\dfrac{1}{8}+\dfrac{5}{8}\) b) \(\dfrac{1}{15}+\dfrac{4}{15}\) c) \(\dfrac{5}{9}+\dfrac{7}{9}\) d) \(\dfrac{23}{100}+\dfrac{27}{100}\)
a: \(\dfrac{1}{8}+\dfrac{5}{8}=\dfrac{1+5}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
b: \(\dfrac{1}{15}+\dfrac{4}{15}=\dfrac{1+4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{5}{9}+\dfrac{7}{9}=\dfrac{5+7}{9}=\dfrac{12}{9}=\dfrac{4}{3}\)
d: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
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Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
SGK Kết nối tri thức với cuộc sống - Trang 81
24 tháng 8 2023 lúc 0:54
Tính (theo mẫu).Mẫu: dfrac{1}{2}-dfrac{5}{12}dfrac{6}{12}-dfrac{5}{12}dfrac{6-5}{12}dfrac{1}{12}a) dfrac{3}{4}-dfrac{1}{8} b) dfrac{2}{6}-dfrac{5}{18} c) dfrac{2}{5}-dfrac{3}{20}
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Tính (theo mẫu).
| Mẫu: \(\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}\) |
a) \(\dfrac{3}{4}-\dfrac{1}{8}\) b) \(\dfrac{2}{6}-\dfrac{5}{18}\) c) \(\dfrac{2}{5}-\dfrac{3}{20}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
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