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Bài giải
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{394}{90}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{3}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\text{ x }\frac{4}{3}\text{ : }\frac{8}{5}=\frac{9}{12}\text{ x }\frac{4}{3}\text{ x }\frac{5}{8}=\frac{9\text{ x }4\text{ x }5}{12\text{ x }3\text{ x }8}=\frac{5}{8}\)
\(\frac{4}{5}\text{ x }\frac{15}{8}\text{ : }\frac{5}{7}=\frac{4}{5}\text{ x }\frac{15}{8}\text{ x }\frac{7}{5}=\frac{4\text{ x }15\text{ x }7}{5\text{ x }8\text{ x }5}=\frac{21}{10}\)
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{197}{45}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{1}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\times\frac{4}{3}:\frac{8}{5}=1:\frac{8}{5}=\frac{5}{8}\)
\(\frac{4}{5}\times\frac{15}{8}:\frac{5}{7}=\frac{3}{2}:\frac{5}{7}=\frac{21}{10}\)
3/8
13
Đặt tử ra ngoài rồi nhóm các phân số có mẫu lần lượt là 7,5,11,37 và tử là 1 vào xong rút gọn đi là ok = 3/4
a) (x - 1) + (x + 2) + (x - 3) + (x + 4) + (x - 5) + (x + 6) = 21
x - 1 + x + 2 + x - 3 + x + 4 + x - 5 + x + 6 = 21
6x + 3 = 21
6x = 21 - 3
6x = 18
x = 18 : 6
x = 3
b) (x - 1) + (x + 3) + (x - 5) + (x + 7) + (x - 9) + (x + 11) = 186
x - 1 + x + 3 + x - 5 + x + 7 + x - 9 + x + 11 = 186
6x + 6 = 186
6x = 186 - 6
6x = 180
x = 180 : 6
x = 30
1.(143 x 2012 + 30 x 4024 - 3 x 2012) : (20 : 10/11 + 23 1/7 x 7/9)
= 1 .[143 x 2012 + 60 x 2012 - 3 x 2012) : (22 + 18)
= [2012 x (143 + 60 - 3)] : 40
= [2012 x 200] : 40
= 2012 x 5
= 10060
Ủng hộ nha
\(\frac{22}{7}:\left(11-x\right)+\frac{2}{3}=\frac{7}{5}\)
\(\frac{22}{7}:\left(11-x\right)=\frac{11}{15}\)
\(11-x=\frac{30}{7}\)
\(x=\frac{47}{7}\)
Vậy \(x=\frac{47}{7}\)
\(\frac{22}{7}\div\left(11-x\right)+\frac{2}{3}=\frac{7}{5}\)
\(\Leftrightarrow\frac{2}{7}-\frac{22}{7x}+\frac{2}{3}=\frac{7}{5}\)
\(\Rightarrow\frac{30x}{105x}+\frac{330}{105x}+\frac{70x}{105x}=\frac{147}{105x}\)
\(\Rightarrow100x+330=147\)
\(\Rightarrow100x=-153\)
\(\Rightarrow x=-1,53\)