các bạn ơi giúp mình giải bài này với:
cho A= 4040/7x9+4040/9x11+4040/1x13+4040/13x15 và B= 2/63+ 2/108+3/180
tính A:B
K
Khách
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Những câu hỏi liên quan
21 tháng 1
Bài 2:
a: \(\dfrac{1}{4};\dfrac{2}{5}\)
\(\dfrac{1}{4}=\dfrac{1\cdot5}{4\cdot5}=\dfrac{5}{20}\)
\(\dfrac{2}{5}=\dfrac{2\cdot4}{5\cdot4}=\dfrac{8}{20}\)
\(\dfrac{2}{3};\dfrac{7}{8}\)
\(\dfrac{2}{3}=\dfrac{2\cdot8}{3\cdot8}=\dfrac{16}{24}\)
\(\dfrac{7}{8}=\dfrac{7\cdot3}{8\cdot3}=\dfrac{21}{24}\)
\(\dfrac{3}{4};\dfrac{5}{6}\)
\(\dfrac{3}{4}=\dfrac{3\cdot3}{4\cdot3}=\dfrac{9}{12}\)
\(\dfrac{5}{6}=\dfrac{5\cdot2}{6\cdot2}=\dfrac{10}{12}\)
b: \(\dfrac{1}{3};\dfrac{7}{9}\)
\(\dfrac{1}{3}=\dfrac{1\cdot3}{3\cdot3}=\dfrac{3}{9}\)
\(\dfrac{7}{9}=\dfrac{7\cdot1}{9\cdot1}=\dfrac{7}{9}\)
\(\dfrac{3}{4};\dfrac{9}{24}\)
\(\dfrac{3}{4}=\dfrac{3\cdot6}{4\cdot6}=\dfrac{18}{24}\)
\(\dfrac{9}{24}=\dfrac{9\cdot1}{24\cdot1}=\dfrac{9}{24}\)
\(\dfrac{7}{10};\dfrac{19}{30}\)
\(\dfrac{7}{10}=\dfrac{7\cdot3}{10\cdot3}=\dfrac{21}{30}\)
\(\dfrac{19}{30}=\dfrac{19\cdot1}{30\cdot1}=\dfrac{19}{30}\)
Bài 1:
\(\dfrac{36}{108}=\dfrac{36:36}{108:36}=\dfrac{1}{3}\)
\(\dfrac{28}{30}=\dfrac{28:2}{30:2}=\dfrac{14}{15}\)
\(\dfrac{42}{98}=\dfrac{42:14}{98:14}=\dfrac{3}{7}\)
\(\dfrac{15}{120}=\dfrac{15:15}{120:15}=\dfrac{1}{8}\)
\(\dfrac{84}{364}=\dfrac{84:28}{364:28}=\dfrac{3}{13}\)
\(\dfrac{120}{100}=\dfrac{120:20}{100:20}=\dfrac{6}{5}\)
\(\dfrac{418}{38}=\dfrac{418:38}{38:38}=\dfrac{11}{1}=11\)
\(\dfrac{96}{1056}=\dfrac{96:96}{1056:96}=\dfrac{1}{11}\)
\(\dfrac{3838}{4040}=\dfrac{3838:101}{4040:101}=\dfrac{38}{40}=\dfrac{38:2}{40:2}=\dfrac{19}{20}\)
\(\dfrac{119119}{123123}=\dfrac{119119:1001}{123123:1001}=\dfrac{119}{123}\)
24 tháng 10 2021
\(2021^2-2021.4040+2020^2=2021^2-2.2021.2020+2020^2\)
\(=\left(2021-2020\right)^2=1^2=1\)
1 tháng 11 2021
\(A=y-2y^2+4040=-2\left(y^2-\dfrac{y}{2}+\dfrac{1}{16}\right)+\dfrac{32321}{8}\)
\(=-2\left(y-\dfrac{1}{4}\right)^2+\dfrac{32321}{8}\le\dfrac{32321}{8}\)
\(maxA=\dfrac{32321}{8}\Leftrightarrow y=\dfrac{1}{4}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
30 tháng 3 2023
Tính nhanh
2020×28+74×2020-4040
= 2020 x 28 + 74 x 2020 - 2020 x 2
= 2020 x (28+74 - 2)
= 2020 x 100
= 202 000
MN
3
29 tháng 8 2023
\(A=\dfrac{2024x2022-4048}{2020x2024+4040}\)
\(A=\dfrac{2024x2022-2x2024}{2020x2024+2x2020}\)
\(A=\dfrac{2024x\left(2022-2\right)}{2020x\left(2024+2\right)}\)
\(A=\dfrac{2024x2020}{2020x2026}\)
\(A=\dfrac{2024}{2026}\)
\(A=\dfrac{1012}{1013}\)
F
6 tháng 2 2020
Bài giải
\(-3x\left|2020x+4040\right|=15\)
\(\left|2020x+4040\right|=15\text{ : }\left(-3x\right)\)
\(\left|2020\left(x+2\right)\right|=-5x\)
\(\Rightarrow\orbr{\begin{cases}2020\left(x+2\right)=5x\\2020\left(x+2\right)=-5x\end{cases}}\Rightarrow\orbr{\begin{cases}x+2=\frac{x}{404}\\x+2=\frac{x}{-404}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{x}{404}-x=x\left(\frac{1}{404}-1\right)=2\\\frac{x}{-404}-x=x\left(-\frac{1}{404}-1\right)=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\cdot\frac{-403}{404}=2\\x\cdot\frac{-405}{404}=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-808}{403}\\x=\frac{-808}{405}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{-808}{403}\text{ ; }\frac{-808}{405}\right\}\)
5 tháng 7 2015
2404+540:(x-6)=4040
540:(x-6)=4040-2404
540:(x-6)=1636
x-6=540:1636
x-6=\(\frac{135}{409}\)
x=\(\frac{135}{409}\)+6
x=\(6\frac{135}{409}\)
tick dung nha
Sửa đề: \(A=\dfrac{4040}{7\times9}+\dfrac{4040}{9\times11}+\dfrac{4040}{11\times13}+\dfrac{4040}{13\times15}\)
\(=2020\left(\dfrac{2}{7\times9}+\dfrac{2}{9\times11}+\dfrac{2}{11\times13}+\dfrac{2}{13\times15}\right)\)
\(=2020\left(\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(=2020\left(\dfrac{1}{7}-\dfrac{1}{15}\right)=2020\times\dfrac{8}{105}\)
Sửa đề: \(B=\dfrac{2}{63}+\dfrac{3}{108}+\dfrac{3}{180}\)
\(=\dfrac{2}{7\times9}+\dfrac{3}{9\times12}+\dfrac{3}{12\times15}\)
\(=\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{15}=\dfrac{1}{7}-\dfrac{1}{15}=\dfrac{8}{105}\)
=>A/B=2020