a)-246 - (-301)
b)-1 1/9 - (-5/12)
c)-3/26 + 24/69
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN MỌI NGƯỜI
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
6
1
10 tháng 8 2023
a) \(0,6+\dfrac{2}{3}=\dfrac{6}{10}+\dfrac{2}{3}=\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{9}{15}+\dfrac{10}{15}=\dfrac{19}{15}\)
b) \(-\dfrac{5}{12}+0,75=-\dfrac{5}{12}+\dfrac{75}{100}=-\dfrac{5}{12}+\dfrac{3}{4}=-\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
c) \(\dfrac{1}{3}-\left(-0,4\right)=\dfrac{1}{3}+\dfrac{4}{10}=\dfrac{1}{3}+\dfrac{2}{5}=\dfrac{5}{15}+\dfrac{6}{15}=\dfrac{11}{15}\)
d) \(1\dfrac{3}{5}+\dfrac{5}{6}=\dfrac{8}{5}+\dfrac{5}{6}=\dfrac{48}{40}+\dfrac{25}{30}=\dfrac{73}{30}\)
NX
17 tháng 8 2023
a) \(\dfrac{3}{8}+\dfrac{15}{-25}+\dfrac{3}{5}\)
\(=\dfrac{-9}{40}+\dfrac{3}{5}\)
\(=\dfrac{3}{8}\)
b) \(\dfrac{-5}{18}+\dfrac{23}{45}-\dfrac{9}{10}\)
\(=\dfrac{7}{30}-\dfrac{9}{10}\)
\(=\dfrac{-2}{3}\)
c) \(\dfrac{-5}{12}+\dfrac{15}{18}-2,25\)
\(=\dfrac{5}{12}-2,25\)
\(=\dfrac{-11}{6}\)
d) \(\dfrac{5}{6}+\dfrac{2}{3}-0,5\)
\(=\dfrac{3}{2}-0,5\)
\(=1\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 8 2023
\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)
NX
16 tháng 8 2023
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
NX
16 tháng 8 2023
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
6
1
31 tháng 7 2023
\([\)6+(\(\dfrac{1}{2}\))3\(]\):\(\dfrac{3}{12}\)=\([\)6+\(\dfrac{1}{8}\)\(]\):\(\dfrac{1}{4}\)=\(\dfrac{49}{8}\):\(\dfrac{1}{4}\)=\(\dfrac{49}{2}\).
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 6 2023
a, 2\(xy\) - 2\(x\) + 3\(y\) = -9
(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12
2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12
(\(y-1\))(2\(x\) + 3) = -12
Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}
Lập bảng ta có:
| \(y\)-1 | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| \(y\) | -11 | -5 | -3 | -2 | -1 | 0 | 2 | 3 | 4 | 5 | 7 | 13 |
| 2\(x\)+3 | 1 | 2 | 3 | 4 | 6 | 12 | -12 | -6 | -4 | -3 | -2 | -1 |
| \(x\) | -1 | -\(\dfrac{1}{2}\) | 0 | \(\dfrac{1}{2}\) | \(\dfrac{3}{2}\) | \(\dfrac{9}{2}\) | \(-\dfrac{15}{2}\) | \(-\dfrac{9}{2}\) | -\(\dfrac{7}{2}\) | -3 | \(-\dfrac{5}{2}\) | -2 |
Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 6 2023
b, (\(x+1\))2(\(y\) - 3) = -4
Ư(4) = {-4; -2; -1; 1; 2; 4}
Lập bảng ta có:
| \(\left(x+1\right)^2\) | - 4(loại) | -2(loại) | -1(loại) | 1 | 2 | 4 |
| \(x\) | 0 | \(\pm\)\(\sqrt{2}\)(loại) | 1; -3 | |||
| \(y-3\) | 1 | 2 | 4 | -4 | -2 | -1 |
| \(y\) | -1 | 2 |
Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (0; -1); (-3; 2); (1; 2)
NX
23 tháng 8 2023
a)\(\dfrac{-5}{21}:\dfrac{25}{14}=\dfrac{-5}{21}.\dfrac{14}{25}=\dfrac{-2}{15}\)
b)\(3,4:\dfrac{-17}{14}=\dfrac{17}{15}:\dfrac{-17}{14}=\dfrac{17}{15}.\dfrac{14}{-17}=\dfrac{-14}{15}\)
c) \(\left(-1,7\right):1\dfrac{2}{15}=\dfrac{-17}{10}:\dfrac{17}{15}=\dfrac{-17}{10}.\dfrac{15}{17}=\dfrac{-3}{2}\)
d) \(-8,4:\left(-2,8\right)=\dfrac{-42}{5}:\dfrac{-14}{5}=\dfrac{-42}{5}.\dfrac{5}{-14}=3\)
a) \(...=-246+301=55\)
b) \(...=-\dfrac{10}{9}+\dfrac{5}{12}=-\dfrac{40}{36}+\dfrac{15}{36}=-\dfrac{25}{36}\)
c) \(...=-\dfrac{3}{26}+\dfrac{24}{69}=-\dfrac{207}{1794}+\dfrac{624}{1794}=\dfrac{417}{1794}\)