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E
5
25 tháng 3 2017
\(\frac{9}{15}x\frac{12}{1}x\frac{5}{6}\)=\(\frac{9x12x5}{15x1x6}\)=\(\frac{9x6x2x5}{5x3x1x6}\)= \(\frac{9x2}{3x1}\)=\(\frac{18}{3}\)= 6
Toán này đâu có phải toán lớp 1? Mà toán này là toán lớp 5.
CN
11
10 tháng 8 2017
a) \(\frac{2}{9}+\frac{1}{5}+\frac{7}{9}+\frac{4}{5}=\left(\frac{2}{9}+\frac{7}{9}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)= 1+1=2 \)
b) \(\frac{19}{37}+\left(1+\frac{19}{37}\right)=\left(\frac{19}{37}-\frac{19}{37}\right)+1=0+1=1\)
c) \(\frac{9}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)+\frac{7}{8}=\frac{9}{8}+\frac{7}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)=2-2=0\)
d) \(\frac{3}{5}\times\frac{8}{27}\times\frac{5}{3}=\frac{3}{5}\times\frac{5}{3}\times\frac{8}{27}=1\times\frac{8}{27}=\frac{8}{27}\)
e) \(\frac{7}{19}\times\frac{1}{3}+\frac{7}{19}\times\frac{2}{3}=\frac{7}{19}\times\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}\times1=\frac{7}{19}\)
f) \(\frac{12}{5}\times4-4\times\frac{7}{5}=4\times\left(\frac{12}{5}-\frac{7}{5}\right)=4\times1=4\)
KU
7
20 tháng 2 2016
số kẹo còn lại là
12 - 11 = 1 ( cái )
đáp số 1 cái kẹo
đúng 100%
duyệt nha
VT
4
25 tháng 4 2016
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{4}{6}+\frac{9}{12}+\frac{16}{20}=\left(\frac{1}{3}+\frac{4}{6}\right)+\left(\frac{1}{4}+\frac{9}{12}\right)+\left(\frac{1}{5}+\frac{16}{20}\right)\)
\(=1+1+1=3\)
HT
25 tháng 4 2016
Ta có\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{4}{6}+\frac{9}{12}+\frac{16}{20}=\left(\frac{1}{3}+\frac{4}{6}\right)+\left(\frac{1}{4}+\frac{9}{12}\right)+\left(\frac{1}{5}+\frac{16}{20}\right)\)
\(=\left(\frac{2+4}{6}\right)+\left(\frac{3+9}{12}\right)+\left(\frac{4+16}{20}\right)\)
\(=\frac{6}{6}+\frac{12}{12}+\frac{20}{20}\)
\(=1+1+1\)
\(=3\)
CC
16
21
12 + 9 = 21
Bằng 21
21
21
21
21
- Tính tổng số phần: 7 + 9 = 16 phần.
- Tìm tổng khối lượng của hai bao: 33kg + 41kg = 74kg.
- Tính khối lượng mỗi bao sau khi giảm:
- Bao thứ nhất: 74kg * (7/16) = 32.5 kg.
- Bao thứ hai: 74kg * (9/16) = 41.25 kg.
Kết quả này vẫn không thể đạt được vì chúng ta chỉ có thể bớt đi. Có thể đề bài có sự nhầm lẫn trong tỉ lệ, hoặc chỉ bớt ở bao thứ hai. Nếu đề bài là bớt ở bao thứ hai để tỉ lệ là 79:- Tính hiệu số lượng bao thứ nhất: 33kg * (7/9) = 38.88...kg
- Tính tỉ lệ 79: 33 * (9/7) = 42.85...kg
Nếu bớt mỗi bao một lượng bằng nhau để bao thứ nhất bằng 7/9 bao thứ hai:- Gọi số kg cần bớt ở mỗi bao là x.
- Ta có phương trình: 33 - x = (7/9) * (41 - x).
- Giải phương trình: 9 * (33 - x) = 7 * (41 - x).
- 297 - 9x = 287 - 7x.
- 10 = 2x => x = 5 kg.
Sau khi bớt 5kg, bao thứ nhất còn 28kg và bao thứ hai còn 36kg.Kiểm tra tỉ lệ: 28 / 36 = 7 / 9. Vậy, bạn cần bớt mỗi bao 5kg để bao thứ nhất bằng 7/9 bao thứ hai.
12+9=21
12+9=21
21
21
12 + 9 = 21
21
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21 de
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21 nhá em
21 chứ nhiêu