Giúp mình với, mình đang cần gấp:
a,4/5+(-1)/9-7/3 ;b,-2/19+8/13-17/19+1 và 3/4+-5/-13; c,9/7-2/7:0,25+2/3; d,4/13*5/11+4/13*6/11-4/13; e,10/17+-5/13+7/17-8/13
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a: \(\dfrac{4}{5}+\dfrac{-1}{9}-\dfrac{7}{3}=\dfrac{4}{5}-\dfrac{1}{9}-\dfrac{21}{9}=\dfrac{4}{5}-\dfrac{20}{9}\)
\(=\dfrac{36}{45}-\dfrac{100}{45}=-\dfrac{64}{45}\)
b: \(-\dfrac{2}{19}+\dfrac{8}{13}-\dfrac{17}{19}+1\dfrac{3}{4}+\dfrac{-5}{-13}\)
\(=\left(-\dfrac{2}{19}-\dfrac{17}{19}\right)+\left(\dfrac{8}{13}+\dfrac{5}{13}\right)+\dfrac{7}{4}\)
\(=-\dfrac{19}{19}+\dfrac{13}{13}+\dfrac{7}{4}=\dfrac{7}{4}\)
c: \(\dfrac{9}{7}-\dfrac{2}{7}:0,25+\dfrac{2}{3}\)
\(=\dfrac{9}{7}-\dfrac{2}{7}:\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{9}{7}-\dfrac{8}{7}+\dfrac{2}{3}\)
\(=\dfrac{1}{7}+\dfrac{2}{3}=\dfrac{3}{21}+\dfrac{14}{21}=\dfrac{17}{21}\)
d: \(\dfrac{4}{13}\cdot\dfrac{5}{11}+\dfrac{4}{13}\cdot\dfrac{6}{11}-\dfrac{4}{13}\)
\(=\dfrac{4}{13}\left(\dfrac{5}{11}+\dfrac{6}{11}-1\right)\)
\(=\dfrac{4}{13}\left(\dfrac{11}{11}-1\right)=\dfrac{4}{13}\cdot0=0\)
e: \(\dfrac{10}{17}+\dfrac{-5}{13}+\dfrac{7}{17}-\dfrac{8}{13}\)
\(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\)
\(=\dfrac{17}{17}-\dfrac{13}{13}=1-1=0\)
Hai phân số 4/3 và -12/9 không bằng nhau.
Giải thích:
Vậy, 4/3 và -12/9 không bằng nhau vì chúng có dấu khác nhau.
1. Quy đồng mẫu số:
2. Khử mẫu:
3. Phân phối và đơn giản hóa:
4. Giải x:
Kết luận:
\(\dfrac{-6}{12}=\dfrac{x}{8}=\dfrac{-7}{y}=\dfrac{z}{-18}\\ \Leftrightarrow\dfrac{-1}{2}=\dfrac{x}{8}=\dfrac{-7}{y}=\dfrac{z}{-18}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{8\cdot\left(-1\right)}{2}=-4\\y=\dfrac{\left(-7\right)\cdot2}{-1}=14\\z=\dfrac{\left(-18\right)\cdot\left(-1\right)}{2}=9\end{matrix}\right.\)
vậy x = -4; y = 14; z = 9
Bước 1: Tìm x
Bước 2: Tìm y
Bước 3: Tìm z
Kết luận:
Để tính tổng của dãy số A = 1/1^2 + 1/2^3 + 1/3^4 + ... + 1/2024^2025, ta có thể nhận thấy rằng đây là một dãy số vô hạn có các số hạng giảm dần rất nhanh.
Phân tích:
Đánh giá:
Kết luận:
\(\dfrac{15}{x}=\dfrac{y}{4}=\dfrac{z}{16}=\dfrac{6}{-8}\\ \Leftrightarrow\dfrac{15}{x}=\dfrac{y}{4}=\dfrac{z}{16}=\dfrac{-3}{4}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot15}{-3}=-20\\y=\dfrac{\left(-3\right)\cdot4}{4}=-3\\z=\dfrac{\left(-3\right)\cdot16}{4}=-12\end{matrix}\right.\\ \text{vậy }x=-20;y=-3;z=-12\)
Đề bài:
15/x = y/4 = z/16 = 6/-8
Cách giải:
1. Tìm giá trị của các tỉ số:
2. Tìm giá trị của x:
3. Tìm giá trị của y:
4. Tìm giá trị của z:
Kết luận:
\(A=\dfrac{3^2}{20\cdot23}+\dfrac{3^2}{23\cdot26}+...+\dfrac{3^2}{77\cdot80}\\ A=\dfrac{3^2}{3}\cdot\left[\left(\dfrac{1}{20}-\dfrac{1}{23}\right)+\left(\dfrac{1}{23}-\dfrac{1}{26}\right)+...+\left(\dfrac{1}{77}-\dfrac{1}{80}\right)\right]\\ A=3\cdot\left[\dfrac{1}{20}-\dfrac{1}{80}\right]\\ A=3\cdot\dfrac{3}{80}=\dfrac{9}{80}< 1\)
\(A=\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+...+\dfrac{3^2}{77.80}\)
\(3A=3^2.\left(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}\right)\)
\(3A=9.\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)\)
\(3A=9\left(\dfrac{1}{20}-\dfrac{1}{80}\right)\)
\(3A=9\left(\dfrac{4}{80}-\dfrac{1}{80}\right)\)
\(3A=9.\dfrac{3}{80}\)
\(3A=\dfrac{27}{80}\)
\(A=\dfrac{27}{80}:3\)
\(A=\dfrac{27}{80}.\dfrac{1}{3}\)
\(A=\dfrac{9}{80}\)
Ta có: \(\dfrac{9}{80}< 1\)
\(\Rightarrow A< 1\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{4950}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+...+\dfrac{2}{9900}\)
\(A=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9900}\right)\)
\(A=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\right)\)
\(A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=2\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(A=2\left(\dfrac{50}{100}-\dfrac{1}{100}\right)\)
\(A=2.\dfrac{49}{100}\)
\(A=\dfrac{49}{50}\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\\ A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{9900}\\ A=2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\\ A=2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\right)\\ A=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ A=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ A=2\cdot\dfrac{49}{100}\\ A=\dfrac{49}{50}\)
a: \(\dfrac{4}{5}+\dfrac{-1}{9}-\dfrac{7}{3}=\dfrac{4}{5}-\dfrac{1}{9}-\dfrac{21}{9}=\dfrac{4}{5}-\dfrac{20}{9}\)
\(=\dfrac{36}{45}-\dfrac{100}{45}=-\dfrac{64}{45}\)
b: \(-\dfrac{2}{19}+\dfrac{8}{13}-\dfrac{17}{19}+1\dfrac{3}{4}+\dfrac{-5}{-13}\)
\(=\left(-\dfrac{2}{19}-\dfrac{17}{19}\right)+\left(\dfrac{8}{13}+\dfrac{5}{13}\right)+\dfrac{7}{4}\)
\(=-\dfrac{19}{19}+\dfrac{13}{13}+\dfrac{7}{4}=\dfrac{7}{4}\)
c: \(\dfrac{9}{7}-\dfrac{2}{7}:0,25+\dfrac{2}{3}\)
\(=\dfrac{9}{7}-\dfrac{2}{7}:\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{9}{7}-\dfrac{8}{7}+\dfrac{2}{3}\)
\(=\dfrac{1}{7}+\dfrac{2}{3}=\dfrac{3}{21}+\dfrac{14}{21}=\dfrac{17}{21}\)
d: \(\dfrac{4}{13}\cdot\dfrac{5}{11}+\dfrac{4}{13}\cdot\dfrac{6}{11}-\dfrac{4}{13}\)
\(=\dfrac{4}{13}\left(\dfrac{5}{11}+\dfrac{6}{11}-1\right)\)
\(=\dfrac{4}{13}\left(\dfrac{11}{11}-1\right)=\dfrac{4}{13}\cdot0=0\)
e: \(\dfrac{10}{17}+\dfrac{-5}{13}+\dfrac{7}{17}-\dfrac{8}{13}\)
\(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\)
\(=\dfrac{17}{17}-\dfrac{13}{13}=1-1=0\)