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\(a,11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{27}{6}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{14}{6}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{7}{3}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{7}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}-\dfrac{12}{3}\)
\(=\dfrac{47}{4}-4\)
\(=\dfrac{47}{4}-\dfrac{16}{4}\)
\(=\dfrac{31}{4}\)
c) Ta có: \(4\dfrac{3}{7}:\left(\dfrac{7}{5}\cdot4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
a: \(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(=\dfrac{11}{2}\cdot\dfrac{15}{4}=\dfrac{165}{8}\)
c: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}-6-\dfrac{2}{9}=6+\dfrac{3}{5}=\dfrac{33}{5}\)
d: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)
a, \(\Leftrightarrow2x^2=72\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm6\)
Vậy ...
\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)
\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)
Vậy ...
\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)
Vậy ...
\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)
\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)
Vậy ...
a) 2x2 - 72 = 0
\(\Rightarrow\) 2x2 = 72
\(\Rightarrow\) x2 = 36 = 62 = (- 6)2
\(\Rightarrow\) x = 6 hoặc x = - 6
Vậy x = 6 hoặc x = - 6
b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)
\(\Rightarrow\) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)
\(\Rightarrow\) x = \(\dfrac{13}{4}\)
Vậy x = \(\dfrac{13}{4}\)
\(C=\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
c\(C=\left(2\frac{2}{6}+3\frac{3}{6}\right):\left(-4\frac{7}{42}+3\frac{6}{42}\right)+7\frac{1}{2}\)
\(C=5\frac{5}{6}:-1\frac{13}{42}+7\frac{1}{2}\)
\(C=\frac{35}{6}:\frac{55}{42}+7\frac{1}{2}\)
\(C=\frac{35}{6}\cdot\frac{42}{55}+7\frac{1}{2}\)
\(C=\frac{49}{11}+\frac{15}{2}\)
\(C=\frac{98}{22}+\frac{165}{22}\)
\(C=\frac{263}{22}\)
k mik nha Nữ Hoàng Giấu Tên
\(a)\)\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}\)
\(=\left(6\frac{4}{5}-3\frac{4}{5}\right)-1\frac{2}{3}\)
\(=\left(6+\frac{4}{5}-3-\frac{4}{5}\right)-\frac{5}{3}\)
\(=3-\frac{5}{3}\)
\(=\frac{9}{3}-\frac{5}{3}\)
\(=\frac{4}{3}\)
\(b)\)\(6\frac{5}{7}-(1\frac{3}{4}+2\frac{5}{7})\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=\left(6+\frac{5}{7}-2-\frac{5}{7}\right)-\frac{7}{4}\)
\(=4-\frac{7}{4}\)
\(=\frac{16}{4}-\frac{7}{4}\)
\(=\frac{9}{4}\)
\(c)\)\(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=\left(7\frac{5}{9}-3\frac{5}{9}\right)-2\frac{3}{4}\)
\(=\left(7+\frac{5}{9}-3-\frac{5}{9}\right)-\frac{11}{4}\)
\(=4-\frac{11}{4}\)
\(=\frac{16}{4}-\frac{11}{4}\)
\(=\frac{5}{4}\)
P/s: Câu b bạn thiếu ngoặc.
Linz
a) 6 và 4/5 - ( 1 và 2/3 + 3 và 4/5)
= 34/5- ( 5/3 + 19/5 )
= 34/5 - 5/3 - 19/5
= (34/5 - 9/5 ) - 5/3
=25/5 -5/3
= 5/1 - 5/3
= 15/3 - 5/3
=10/3
b) 6 và 5/7 - (1 và 3/4 + 2 và 5/7)
=47/7 - (7/4 + 19/7 )
= 47/7 -7/4 - 19/7
= (47/7 -19/7 ) - 7/4
= 28/7 -7/4
=4/1 - 7/4
= 16/4 - 7/4
= 9/4
c) 7 và 5/9 - ( 2 và 3/4 + 3 và 5/9 )
= 68/9 - ( 11/4 + 32/9 )
= 68/9 - 11/4 - 32/9
= (68/9 - 32/9 )- 11/4
= 36/9 -11/4
= 4/1 - 11/4
= 16/4 - 11/4
= 5/4
Chúc bạn học tốt nha!!!!!
a. \(\dfrac{-6}{15}=\dfrac{-6:3}{15:3}=\dfrac{-2}{5}\)
Vậy \(\dfrac{-6}{15}\ne\dfrac{2}{-3}\)
b. \(\dfrac{-6}{-7}=\dfrac{-6.\left(-1\right)}{-7.\left(-1\right)}=\dfrac{6}{7}\)
Vậy: \(\dfrac{6}{7}=\dfrac{-6}{-7}\)
c. Ta có: \(\dfrac{5}{7}< 1\)
\(\dfrac{7}{5}>1\)
Vậy \(\dfrac{5}{7}\ne\dfrac{7}{5}\)
a: \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)\)
\(=\dfrac{37}{3}-3-4-\dfrac{3}{2}\)
\(=\dfrac{74-9}{6}-7=\dfrac{65}{6}-7=\dfrac{65-42}{7}=\dfrac{23}{7}\)
b: \(3\dfrac{5}{6}+2\dfrac{1}{6}\cdot6\)
\(=3+\dfrac{5}{6}+\dfrac{13}{6}\cdot6\)
\(=16+\dfrac{5}{6}=\dfrac{101}{6}\)
c: \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
\(=3+\dfrac{1}{2}+4+\dfrac{5}{7}-5-\dfrac{5}{14}\)
\(=2+\dfrac{7+10-5}{14}=2+\dfrac{12}{14}=2+\dfrac{6}{7}=\dfrac{20}{7}\)
d: \(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99+2}{22}=\dfrac{101}{22}\)