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7/3:0.2x=7/6:0.8
7/3 : 0.2 x = 35/24
0.2 x = 35/24 : 7/3
0.2 x = 5/8
x=5/8 : 0.2
x= 25/8
Vậy x=25/8
=>\(\dfrac{7}{3}:0,2x=\dfrac{7}{6}.\dfrac{5}{4}\)
=>\(\dfrac{7}{3}:0,2x=\dfrac{35}{24}\)
=>0,2\(.x=\dfrac{7}{3}:\dfrac{35}{24}\)
=>0,2\(.x=\dfrac{7}{3}.\dfrac{24}{35}\)
=>0,2.\(x=\dfrac{8}{5}\)
=>\(x=\dfrac{8}{5}:0,2\)
=>\(x=\dfrac{8}{5}.5\)
=>\(x=8\)
Vậy x=8
\(\dfrac{7}{3}:0,2x=\dfrac{7}{6}:0,8\)
⇔ \(\dfrac{7}{3}:0,2x=\dfrac{35}{24}\)
⇔ \(0,2x=\dfrac{7}{3}:\dfrac{35}{24}\)
⇔ \(0,2x=1,6\)
⇔ x = 1,6 : 0,2
⇔ x = 8
Vậy x =8
Đây là tính hợp lí ... mà câu a là 27,5 chứ không phải 2,75...
\(A=\dfrac{7,5-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{27,5-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{15}{2}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{55}{2}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{3}}\\ =\dfrac{3\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
b: \(=26:\left[\dfrac{3:0.1}{2.5\cdot2}+\dfrac{0.25\cdot4}{2}\right]+\dfrac{2}{3}\cdot\dfrac{21}{4}\)
\(=26:\left[\dfrac{30}{5}+1\right]+\dfrac{42}{12}\)
\(=\dfrac{26}{7}+\dfrac{42}{12}=\dfrac{101}{14}\)
c: \(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{\left(\dfrac{1}{34}+\dfrac{33}{34}\right)}{\dfrac{1}{2}+\dfrac{9}{2}}=1:5=\dfrac{1}{5}\)
a/Ta có: \(\dfrac{4}{3}-\left[\left(\dfrac{-11}{6}\right)-\left(\dfrac{2}{9}+\dfrac{5}{3}\right)\right]\)
\(=\) \(\dfrac{4}{3}-\left[\dfrac{-11}{6}-\dfrac{2}{9}-\dfrac{5}{3}\right]\)
\(=\) \(\dfrac{4}{3}+\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{3}\)
\(=\) \(\dfrac{24}{18}+\dfrac{33}{18}+\dfrac{4}{18}+\dfrac{30}{18}\)
\(=\) \(\dfrac{91}{18}\)
b/Ta có: \(\left(8-\dfrac{9}{4}+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(3+\dfrac{2}{4}-\dfrac{9}{7}\right)\)
\(=\) \(8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=\) \(8+6-3-\dfrac{9}{4}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\)
\(=\) \(11-\dfrac{2}{4}+\dfrac{14}{7}\)
\(=\) \(11-\dfrac{1}{2}+2\)
\(=\) \(9-\dfrac{1}{2}\)
\(=\) \(\dfrac{17}{2}\)
Chúc bn học tốt!!!
\(a)\dfrac{5}{4}.\dfrac{-12}{7}=\dfrac{5.\left(-12\right)}{4.7}=\dfrac{-60}{28}=\dfrac{-15}{7}\)
\(b)\dfrac{-4}{3}:\dfrac{13}{9}=\dfrac{-4}{3}.\dfrac{9}{13}=\dfrac{\left(-4\right).9}{3.13}=\dfrac{-36}{39}=\dfrac{-12}{13}\)
\(c)\dfrac{-5}{7}.\dfrac{49}{3}:\dfrac{7}{-6}=\dfrac{-5}{7}.\dfrac{49}{3}:\dfrac{-7}{6}=\dfrac{-35}{3}.\dfrac{-6}{7}=10\)
\(d)\left(-\dfrac{9}{25}\right):6=\dfrac{-3}{50}\)
Chúc bn học tốt!
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) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}.\)
\(=\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}.\dfrac{3}{5}-\dfrac{2}{7}.\dfrac{3}{5}.\)
\(=\dfrac{3}{5}\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right).\)
\(=\dfrac{3}{5}.1=\dfrac{3}{5}.\)
Vậy.....
\(\dfrac{3^7.16^3}{12^5.27^2}=\dfrac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}.\)
\(=\dfrac{3^7.2^{12}}{\left(2^2\right)^5.3^5.3^6}=\dfrac{3^7.2^{12}}{2^{10}.3^{11}}.\)
\(=\dfrac{2^2}{3^4}=\dfrac{4}{81}.\)
Vậy.....
7/3:0,2X=7/6:0,8
35/3X=35/24
X=35/24:35/3
X=1/8
7/3 : 0,2x = 7/6 : 0,8
7/3 : 0,2x = 35/24
0,2x = 7/3 : 35/24
0,2x = 8/5
x = 8/5 : 0,2
x = 8