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a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)
b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)
\(\dfrac{18}{3}=18:3=6\)
\(\dfrac{45}{9}=45:9=5\)
\(\dfrac{8}{8}=8:8=1\)
\(\dfrac{7}{1}=7:1=7\)
= 4/5 + 1/5 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
= (4/5 + 1/5) + (1/2 + 1/2) + (1/2 + 1/2) + 1/2
= 1 + 1 + 1 + 1/2
= 3 + 1/2
= 6/2 + 1/2 (Hoặc ra luôn là hỗn số 3 và 1/2)
= 7/2
Ta có: 272727x28 - 282828x27 = 27x10101x28 - 28x10101x27 = 0
Vậy, phép tính đã cho bằng 0
4) \(\frac{29x101-101}{2x19x101+4x101}\)
= \(\frac{\left(29-1\right)x101}{\left(38-4\right)x101}\)
=\(\frac{28}{34}=\frac{14}{17}\)
5) \(\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)-\frac{3}{8}-\frac{5}{8}+\frac{1}{3}\)
= \(\frac{14}{7}-\frac{8}{8}+\frac{1}{3}\)
= \(2-1+\frac{1}{3}=\frac{4}{3}\)
Nhớ tk cho mình nhé
1\(\frac{423133x846267+423134}{423133x846267+423134}=1\))
\(\frac{\left(423133+1\right)x846267-423133}{423133x846267+423134}\)
=\(\frac{423133x846267+846267-423133}{423133x846267+423134}\)
= \(\frac{423133x846267+423134}{423133x846267+423134}\)
= 1
Nhớ tk cho mình nhé (các câu sau làm riêng cho đỡ dài
a) $\frac{1}{6}:\frac{3}{7} = \frac{1}{6} \times \frac{7}{3} = \frac{7}{{18}}$
b) $\frac{5}{{12}}:\frac{1}{4} = \frac{5}{{12}} \times \frac{4}{1} = \frac{{5 \times 4}}{{12 \times 1}} = \frac{{5 \times 4}}{{4 \times 3 \times 1}} = \frac{5}{3}$
c) $\frac{4}{{15}}:\frac{8}{3} = \frac{4}{{15}} \times \frac{3}{8} = \frac{{4 \times 3}}{{15 \times 8}} = \frac{{4 \times 3}}{{5 \times 3 \times 4 \times 2}} = \frac{1}{{10}}$
d) $\frac{{18}}{5}:\frac{9}{{10}} = \frac{{18}}{5} \times \frac{{10}}{9} = \frac{{18 \times 10}}{{5 \times 9}} = \frac{{9 \times 2 \times 5 \times 2}}{{5 \times 9}} = 4$
Đáp án lần lượt là: \(\frac{23}{36};\frac{31}{7};\frac{1}{6};\frac{3}{5}\)
#Hk_tốt
#Ken'z
\(\frac{12}{18}+\frac{1}{3}+\frac{1}{7}+\frac{2}{8}+\frac{27}{36}+\frac{42}{49}\)
\(=\frac{2}{3}+\frac{1}{3}+\frac{1}{7}+\frac{1}{4}+\frac{3}{4}+\frac{6}{7}\)
\(=\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)\)
\(=1+1+1\)
\(=3\)