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B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
\(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(=\frac{11}{125}-\left(\frac{17}{18}-\frac{8}{18}\right)+\left(\frac{17}{14}-\frac{10}{14}\right)\)
\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}+\left(\frac{1}{2}-\frac{1}{2}\right)\)= \(\frac{11}{125}\)
b) \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=\left(6-5-3\right)+\left(\frac{7}{3}-\frac{5}{3}-\frac{2}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0+\frac{-1}{2}\)
= \(-2-\frac{-1}{2}=-\left(2+\frac{1}{2}\right)=-2\frac{1}{2}\)
m) (\(\frac{-5}{12}\)+\(\frac{6}{11}\))+(\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\))
= \(\frac{-5}{12}\)+\(\frac{6}{11}\)+\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\)
= (\(\frac{-5}{12}\)+\(\frac{5}{12}\))+(\(\frac{6}{11}\)+\(\frac{5}{11}\))+\(\frac{7}{17}\)
= 0+1+\(\frac{7}{17}\)
= \(\frac{24}{17}\)
n) (\(\frac{9}{16}\)+\(\frac{8}{-27}\))+(1+\(\frac{7}{16}\)+\(\frac{-19}{27}\))
= \(\frac{9}{16}\)+\(\frac{8}{-27}\)+1+\(\frac{7}{16}\)+\(\frac{-19}{27}\)
= (\(\frac{9}{16}\)+\(\frac{7}{16}\))+(\(\frac{8}{-27}\)+\(\frac{-19}{27}\))+1
= 1+(-1)+1
= 0+1
= 1
o) (6-2\(\frac{4}{5}\)).3\(\frac{1}{8}\)-1\(\frac{3}{5}\):\(\frac{1}{4}\)
= (6-\(\frac{14}{5}\)).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)
= \(\frac{16}{5}\).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)
= 10-\(\frac{8}{5}\):\(\frac{1}{4}\)
= 10-\(\frac{32}{5}\)
= \(\frac{18}{5}\)
CHÚC BẠN HỌC TỐT
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)
a,\(\frac{31}{1000}\)
b,\(\frac{8}{108}\)
c,0
a,\(\frac{49}{97}\)
b,\(\frac{-1}{4751}\)
Ta có:
\(\left(\right. \frac{13 \frac{2}{9} - 15 \frac{2}{3}}{18 \frac{3}{7} - 17 \frac{1}{4}} \cdot \frac{30^{2} - 5^{4}}{25 - 12 \cdot 5^{2}} \left.\right) \cdot x = \frac{\frac{2}{11} + \frac{3}{13} + \frac{4}{15} + \frac{5}{17}}{4 \frac{1}{11} + \frac{5}{13} + \frac{9}{15} + \frac{13}{17}}\)
Bước 1: Đổi hỗn số về phân số
- \(13 \frac{2}{9} = \frac{119}{9}\),
- \(15 \frac{2}{3} = \frac{47}{3}\),
- \(18 \frac{3}{7} = \frac{129}{7}\),
- \(17 \frac{1}{4} = \frac{69}{4}\)
Bước 2: Tính toán từng phần
Ta có:
\(\frac{119}{9} - \frac{47}{3} = \frac{119 - 141}{9} = \frac{- 22}{9}\) \(\frac{129}{7} - \frac{69}{4} = \frac{516 - 483}{28} = \frac{33}{28}\) \(30^{2} - 5^{4} = 900 - 625 = 275\) \(25 - 12 \cdot 25 = 25 - 300 = - 275\)
Khi đó:
\(\left(\right. \frac{- 22}{9} \div \frac{33}{28} \cdot \frac{275}{- 275} \left.\right) = \left(\right. \frac{- 22}{9} \cdot \frac{28}{33} \cdot \left(\right. - 1 \left.\right) \left.\right) = \frac{616}{297}\)
Bước 3: Tính vế phải
Tử số:
\(\frac{2}{11} + \frac{3}{13} + \frac{4}{15} + \frac{5}{17} = \frac{35494}{36465}\)
Mẫu số:
\(4 \frac{1}{11} + \frac{5}{13} + \frac{9}{15} + \frac{13}{17} = \frac{149645}{36465}\)
→ Vế phải:
\(\frac{35494}{36465} \div \frac{149645}{36465} = \frac{35494}{149645}\)
Bước 4: Giải phương trình
\(\frac{616}{297} \cdot x = \frac{35494}{149645} \Rightarrow x = \frac{35494}{149645} \cdot \frac{297}{616} = \frac{813}{7118}\)
Vậy:
\(\boxed{x = \frac{813}{7118}}\)
hỉu không =]]]
2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)
3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
=2/7-2/7=0
\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\) _ \(\frac{3-\frac{3}{11}-\frac{3}{17}}{5-\frac{5}{11}-\frac{5}{17}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)_ \(\frac{3\left(1-\frac{1}{11}-\frac{1}{17}\right)}{5\left(1-\frac{1}{11}-\frac{1}{17}\right)}\)= \(\frac{2}{4}-\frac{3}{5}\)= \(\frac{-1}{10}\)