Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)
\(=\frac{53}{101}.\frac{-97}{97}\)
\(=\frac{53}{101}.\left(-1\right)\)
\(=\frac{-53}{101}\)
b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)
\(=0\)
c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)
\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)
\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)
\(=\frac{3.3.5.2}{\left(-7\right).7}\)
\(=\frac{90}{-49}\)
d) \(\frac{25.48-25.18}{20.5^3}\)
\(=\frac{25\left(48-18\right)}{10.2.125}\)
\(=\frac{25.10.3}{10.2.25.5}\)
\(=\frac{3}{10}\)
\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{101}+\frac{1}{102}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{51}\)
\(=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)
\(=VP\)
câu 1b
Gọi d là ƯCLN (3n-7, 2n-5), d thuộc N*
Ta có : 3n-7 chia ht cho d , 2n_5 chia ht cho d
suy ra: 2(3n-7) chia ht cho d , 3(2n-5) chia ht cho d
suy ra 6n-14 chia ht cho d, 6n-15 chia ht cho d
dấu suy ra [(6n -15) - (6n-14)] chia ht cho d dấu suy ra 1 chia ht cho d suy ra d =1
Vậy......
1) b. Để chứng tỏ \(\frac{3n-7}{2n-5}\) là phân số tối giản
Ta cần chứng minh: ( 3n - 7; 2n - 5 ) = 1
Thật vậy: ( 3n - 7 ; 2n - 5 ) = ( 2n - 5 ; ( 3n - 7 ) - ( 2n - 5 ) ) = ( 2n - 5; n - 2 ) = ( n - 2; n - 3 ) = ( n - 2; 1 ) = 1
=> \(\frac{3n-7}{2n-5}\) là phân số tối giản
3) \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{12}\)
Ta có: \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\left(\frac{1}{5}+\frac{1}{7}\right)+\frac{1}{6}=\frac{12}{35}+\frac{1}{6}>\frac{12}{36}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{1}{2}\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}=\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}\right)>\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \)
=> A > 1/2 + 1/2 + 1/2 + 1/2 = 2
Bg
a)\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1^2.2^2.3^2.....99^2.100^2}{1.2.2.3.3.4.....99.100.100.101}\)
\(=\frac{1^2}{101}\)
\(=\frac{1}{101}\)
Ghi chú: \(=\frac{1^2.2^2.3^2.....99^2.100^2}{1.2.2.3.3.4.....99.100.100.101}\)--> 22 chịt tiêu 2.2 (trên và dưới) làm thế này mãi đến khi còn \(\frac{1^2}{101}\).
b) \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{59^2}{58.60}\)
=\(\frac{2^2.3^2.4^2.....59^2}{1.3.2.4.3.5.....58.60}\)
= \(\frac{2}{1}.\frac{59}{60}\)
= \(\frac{59}{30}\)
Ghi chú: \(\frac{2^2.3^2.4^2.....59^2}{1.3.2.4.3.5.....58.60}\)--> chịt tiêu liên tục, còn \(\frac{2}{1}.\frac{59}{60}\).
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{102}\) (đpcm)
a) \(\frac{53}{101}\cdot-\frac{13}{97}+\frac{53}{101}\cdot-\frac{84}{97}\)
\(=\frac{53}{101}\cdot\left(-\frac{13}{97}-\frac{84}{97}\right)\)
\(=\frac{53}{101}\cdot\left(-1\right)\)
\(=-\frac{53}{101}\)