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

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\\ =\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\\ =\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}\\ =\frac{49}{100}\)

a: \(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{-8+15}{18}:17\)

\(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{7}{18}\cdot\dfrac{1}{17}\)

\(=\dfrac{619}{918}\)

b: \(=\left(3-7+\dfrac{1}{4}\right)\cdot\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{-15}{4}\cdot\dfrac{13}{12}=\dfrac{-195}{48}=\dfrac{-65}{16}\)

c: \(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{2}+\dfrac{6}{5}=\dfrac{6}{5}-1=\dfrac{1}{5}\)

d: \(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{3}{2}\right)\)

\(=\dfrac{5}{16}:\dfrac{-3}{16}=\dfrac{-5}{3}\)

help me !

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Bài này áp dụng quy tắc chuyển vế là ra

\(\left(42-98\right)-42-12\)

\(=\left(42-42\right)-\left(98+12\right)\)

\(=0-110\)

\(=-110\)

(42-98)-42-12

= (42-42)-98+12

=0-110

= -110

Chúc bạn học giỏi !!!!

(42-98)-42-12

= (42-42)-98+12

=0-110

= -110

Chúc bn hc tốt

bn tick mk nha !

bn vaò trang nha Trần Nguênthu

Dạng phân số nha bạn :

Rộng \(\frac{5}{1}\)m và dài\(\frac{8}{1}\)m

Giải

Diện tích căn phòng là:

5/1 . 8/1=40/1=40(m²)

Chu vi căn phòng là:

(5/1 + 8/1).2=36/1=36(m)

Đ/s: diện tích: 40m2

Chu vi 36m

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1999}{2000}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1999}{2000}\)

\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{1999}{2000}:2\)

\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{1999}{4000}\)

\(\dfrac{1}{n+1}=\dfrac{1}{2}-\dfrac{1999}{4000}\)

\(\dfrac{1}{n+1}=\dfrac{1}{4000}\)

\(\Rightarrow n+1=4000\)

\(n=4000-1\)

\(n=3999\)

Vậy n=3999

sửa đề: \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{22.25}\)

giải:

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{22.25}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{22}-\dfrac{1}{25}\\ =1-\dfrac{1}{25}=\dfrac{24}{25}\)

Phần \(\dfrac{1}{22.25}\) đúng đề à?

Giải:

\(\dfrac{-8}{15}< \dfrac{x}{40}\le\dfrac{-7}{15}\)

\(\Leftrightarrow\dfrac{-8.8}{15.8}< \dfrac{x.3}{40.3}\le\dfrac{-7.8}{15.8}\)

\(\Leftrightarrow\dfrac{-64}{120}< \dfrac{3x}{120}\le\dfrac{-56}{120}\)

\(\Leftrightarrow-64< 3x\le-56\)

\(\Leftrightarrow3x\in\left\{-63;-62;-61;-60;-59;-58;-57;-56\right\}\)

\(\Leftrightarrow x\in\left\{-21;-\dfrac{62}{3};-\dfrac{61}{3};-20;-\dfrac{59}{3};-\dfrac{58}{3};-19;-\dfrac{56}{3}\right\}\)

Mà x ∈ Z

\(\Leftrightarrow x\in\left\{-21;-20;-19\right\}\)

Vậy ...

Thay b vào, ta có:\(B=\dfrac{3}{4}.\dfrac{6}{19}+\dfrac{4}{3}.\dfrac{6}{19}-\dfrac{1}{2}.\dfrac{6}{19}=\dfrac{1}{2}\)

Thay c vào, ta có:\(C=\dfrac{2002}{2003}.\dfrac{3}{4}+\dfrac{2002}{2003}.\dfrac{5}{6}-\dfrac{2002}{2003}.\dfrac{19}{12}=0\)