Bài 11: Tính chất cơ bản của phép nhân phân số

Help me pls!

Các dấu * là dấu nhân và các dấu / là dấu phân số.

Bài 1: 

b) Ta có: \(B=\dfrac{\dfrac{4}{17}-\dfrac{4}{49}}{\dfrac{-3}{17}+\dfrac{3}{49}}\)

\(=\dfrac{4\left(\dfrac{1}{17}-\dfrac{1}{49}\right)}{-3\left(\dfrac{1}{17}-\dfrac{1}{49}\right)}=\dfrac{4}{-3}=\dfrac{-4}{3}\)

Giải:

Để A là số nguyên thì:

\(A\in Z\)

\(\Leftrightarrow\dfrac{3}{n-5}\in Z\)

\(\Leftrightarrow3⋮n-5\)

\(\Leftrightarrow n-5\inƯ\left(3\right)\)

\(\Leftrightarrow n-5=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow n=\left\{4;6;2;8\right\}\)

Vậy ...

Để A có giá trị là số nguyên thì

\(\Rightarrow3⋮n-5\)

\(\Rightarrow n-5\inƯ\left(3\right)\)

\(\Rightarrow n-5\in\left\{-1;1;-3;3\right\}\)

Ta có bảng sau:

Vậy \(n\in\left\{4;6;2;8\right\}\)

\(\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}\right):\dfrac{11}{30}\)

=\(\dfrac{21}{4}\)+(\(\dfrac{-25}{30}+\dfrac{3}{30}\)):\(\dfrac{11}{30}\)

=\(\dfrac{21}{4}\)+\(\dfrac{-11}{15}\):\(\dfrac{11}{30}\)

=\(\dfrac{21}{4}\)+(-2)

=\(\dfrac{21}{4}\)+\(\dfrac{-8}{4}\)=\(\dfrac{13}{4}\)

\(\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}\right):\dfrac{11}{30}\)

\(=\dfrac{21}{4}+\dfrac{-11}{15}:\dfrac{11}{30}\)

\(=\dfrac{21}{4}+\left(-2\right)\)

\(=\dfrac{13}{4}\)

Đặt\(B=\)\(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)

\(=5^2.\left(\dfrac{1}{1.6}+\dfrac{1}{6.11}+...+\dfrac{1}{26.31}\right)\)

\(=25.\left(\dfrac{1}{1.6}+\dfrac{1}{6.11}+...+\dfrac{1}{26.31}\right)\)

Đặt \(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+...+\dfrac{1}{26.31}\)

\(\Rightarrow5A=\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\)

\(\Rightarrow5A=\dfrac{6-1}{1.6}+\dfrac{11-6}{6.11}+...+\dfrac{31-26}{26.31}\)

\(\Rightarrow5A=\dfrac{6}{1.6}-\dfrac{1}{1.6}+\dfrac{11}{6.11}-\dfrac{6}{6.11}+...+\dfrac{31}{26.31}-\dfrac{26}{26.31}\)

\(\Rightarrow5A=\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\)

\(\Rightarrow5A=\dfrac{1}{1}-\dfrac{1}{31}\)

\(\Rightarrow5A=\dfrac{31}{31}-\dfrac{1}{31}\)

\(\Rightarrow5A=\dfrac{30}{31}\)

\(\Rightarrow A=\dfrac{30}{31}:5=\dfrac{30}{31}.\dfrac{1}{5}=\dfrac{6}{31}\)

\(\Rightarrow B=25.\dfrac{6}{31}=\) \(\dfrac{150}{31}\) \(=4\dfrac{26}{31}\)

Vậy \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}=4\dfrac{26}{31}\)

a,

\(B=\left(\dfrac{5}{7}\cdot0,6-5:\dfrac{7}{2}\right)\cdot\left(40\%-1,4\right)\cdot\left(-2\right)^3\\ =\left(\dfrac{5}{7}\cdot\dfrac{3}{5}-5\cdot\dfrac{2}{7}\right)\cdot\left(\dfrac{2}{5}-\dfrac{7}{5}\right)\cdot\left(-8\right)\\ =\left(\dfrac{3}{7}-\dfrac{10}{7}\right)\cdot\left(-1\right)\cdot\left(-8\right)\\ =\left(-1\right)\cdot\left(-1\right)\cdot\left(-8\right)\\ =\left(-8\right)\)

b,

\(\dfrac{1}{2}\cdot x+\dfrac{3}{5}\cdot\left(x-2\right)=3\\ \dfrac{1}{2}x+\dfrac{3}{5}x-\dfrac{6}{5}=3\\ x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=3+\dfrac{6}{5}\\ x\cdot\dfrac{11}{10}=\dfrac{21}{5}\\ x=\dfrac{21}{5}:\dfrac{11}{10}\\ x=\dfrac{42}{11}\\ Vậy......\)

- Tham khảo ở đây đi : Câu hỏi của Nguyễn Thị Bích Phương - Toán lớp 6 | Học trực tuyến

Đặt A=\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\)

Vì \(\dfrac{1}{101}\)>\(\dfrac{1}{102}\)>\(\dfrac{1}{103}\)>...>\(\dfrac{1}{300}\)

=>(\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{200}\))+(\(\dfrac{1}{201}\)+\(\dfrac{1}{202}\)+\(\dfrac{1}{203}\)+...+\(\dfrac{1}{300}\)) > (\(\dfrac{1}{200}\)+\(\dfrac{1}{200}\)+\(\dfrac{1}{200}\)+...+\(\dfrac{1}{200}\))+(\(\dfrac{1}{300}\)+\(\dfrac{1}{300}\)+\(\dfrac{1}{300}\)+...+\(\dfrac{1}{300}\)) =>\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) > \(\dfrac{1}{200}\).100 +\(\dfrac{1}{300}\) .100

=> A > \(\dfrac{1}{2}+\dfrac{1}{3}\)

=> A > \(\dfrac{5}{6}\) Mà \(\dfrac{5}{6}\)>\(\dfrac{2}{3}\)=> A > \(\dfrac{2}{3}\) Vậy \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) >\(\dfrac{2}{3}\)

Tra lời:

Ta có:

1/101➢1/300+1/102➢1/300+1/103➢1/300+1/104➢1/300+.....+1/299➢1/300

=1/101+1/102+1/103+...1/299➢199/300

=1/101+1/102+1/103+...1/299+1/300➢199/300+1/300

=200/300=2/3.

Note: ➢ là dau lớn do nhe. Nho tick cho minh nha😊😉

a)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=\dfrac{1}{5}-\dfrac{1}{25}\)

\(=\dfrac{4}{25}\)

b)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)

tương tự

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)