Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)
a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)
\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)
\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)
\(\Rightarrow x=-\dfrac{55}{6}\)
b) 4,25 : 8 = -3,5 : x
\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)
\(x=\dfrac{-3,5.8}{4,25}\)
\(x=\dfrac{-28}{4,25}\)
2.
\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)
\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
Ta có thể lặp đc các tỉ lệ thức sau:
\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)
\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)
\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
2.
a) \(\dfrac{x-1}{4}=\dfrac{9}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=9.4\)
\(\Leftrightarrow\left(x-1\right)^2=36\)
\(\Leftrightarrow\left(x-1\right)^2=\left[{}\begin{matrix}6^2\\-6^2\end{matrix}\right.\)
\(\Leftrightarrow x-1=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}7\\-5\end{matrix}\right.\)
1. a. \(\dfrac{3}{-1}=\dfrac{-15}{5}\); \(\dfrac{-1}{3}=\dfrac{5}{-15}\)
\(\dfrac{3}{-15}\)= \(\dfrac{5}{-1}\); \(\dfrac{-15}{3}=\dfrac{-1}{5}\)
b. \(\dfrac{4}{-3}=\dfrac{-12}{9};\dfrac{-3}{4}=\dfrac{9}{-12}\)
\(\dfrac{4}{-12}=\dfrac{9}{-3};\dfrac{-12}{4}=\dfrac{-3}{9}\)
c. \(\dfrac{3}{7}=\dfrac{c}{b};\dfrac{7}{3}=\dfrac{b}{c}\)
\(\dfrac{3}{c}=\dfrac{b}{7};\dfrac{c}{3}=\dfrac{7}{b}\)
d. \(\dfrac{a}{b}=\dfrac{y}{x};\dfrac{b}{a}=\dfrac{x}{y}\)
\(\dfrac{a}{y}=\dfrac{x}{b};\dfrac{y}{a}=\dfrac{b}{x}\)
chúc bạn học tốt
a) \(\dfrac{1}{3};\) \(\dfrac{2,5}{5,5}=\dfrac{25}{55}=\dfrac{5}{11}\);
\(4:12=\dfrac{4}{12}=\dfrac{1}{3}\) ; \(\dfrac{7}{4}\)
Ta có :\(\dfrac{1}{3}=\dfrac{1}{3}\)\(\Rightarrow\)\(\dfrac{1}{3}=4:12\) nên 2 tỉ số này lập thành 1 tỉ lệ thức.
b) \(\dfrac{4}{9}\); \(\dfrac{18}{42}=\dfrac{3}{7}\); \(\dfrac{-2}{-4,5}=\dfrac{2}{4,5}=\dfrac{20}{45}=\dfrac{4}{9}\);
\(21:49=\dfrac{21}{49}=\dfrac{3}{7}\); \(\dfrac{5}{9}\).
Ta có : - \(\dfrac{4}{9}=\dfrac{4}{9}\Rightarrow\dfrac{4}{9}=\dfrac{-2}{-4,5}\) nên 2 tỉ số này lập thành 1 tỉ lệ thức.
- \(\dfrac{3}{7}=\dfrac{3}{7}\Rightarrow\dfrac{18}{42}=21:49\) nên 2 tỉ số này lập thành 1 tỉ lệ thức.
Chúc bạn hok giỏi nha!
Bài 1:
\(A=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}:\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
\(A=\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}:\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{2}{7}.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\)
\(A=\dfrac{2}{7}:\dfrac{2}{7}=1\)
Bài 2: Here
Chúc bạn học tốt!!!
1. Giải:
Gọi A =M : N
Ta có:M=\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\)= \(\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\)=\(\dfrac{2}{7}\)
N=\(\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)=\(\dfrac{2.\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7.\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\)=\(\dfrac{2}{7}\)
Vậy A=M: N \(\Rightarrow\)A=\(\dfrac{2}{7}\):\(\dfrac{2}{7}\)=\(\dfrac{2}{7}\).\(\dfrac{7}{2}\)=\(\dfrac{2.7}{7.2}\)=1
2. Giải:
Với mọi x \(\in\)Q, ta luôn có \(x\) \(\le\) \(|x|\)(dấu bằng xảy ra khi x\(\ge\)0)
a)Nếu \(x+y\)\(\ge\)0 thì\(|x+y|=x+y\).
Vì \(x\le|x|,y\le|y|\)với mọi x, y\(\in\)Q nên:\(|x+y|=x+y\le|x|+|y|\)
b)Nếu x+y < 0 thì\(|x+y|=-\left(x+y\right)\)=\(-x-y\)
Mà -x\(\le\)\(|x|\), -y\(\le\)\(|y|\) nên: \(|x+y|\)= -x-y\(\le\)\(|x|+|y|\)
Vậy với mọi x, y\(\in\)Q ta đều có:\(|x+y|\le|x|+|y|\). Dấu bằng xảy ra khi x, y cùng dấu hoặc ít nhất có một số bằng 0.
a, Ta có :
\(3:\dfrac{5}{6}=3.\dfrac{6}{5}=\dfrac{5}{2}=2,5\)
\(\dfrac{4}{5}:8=\dfrac{4}{5}.\dfrac{8}{1}=\dfrac{1}{10}=0,1\)
Vì 2,5 \(\ne\) 0,1
=> k thể lập đc thành các TL thức từ các số trên
mk sửa xíu nha
a,Ta có:
\(3:\dfrac{5}{6}=\dfrac{18}{5}=3,6\)
\(\dfrac{4}{5}:8=\dfrac{1}{10}=0,1\)
Vì 3,6 \(\ne\) 0,1
=> k thể lập đc TL thức từ các số trên
Vậy ....
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a) \(\dfrac{-2}{5}=\dfrac{-14}{35}\)
b) \(\dfrac{9}{4}=\dfrac{\dfrac{1}{4}}{\dfrac{1}{9}}\)
c) \(\dfrac{\dfrac{7}{4}}{\dfrac{1}{2}}=\dfrac{7}{2}\)