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Cái này chỉ cần làm quy tắc nhân chéo là ra rồi nhé :)
a) \(x=\dfrac{-2,6.42}{-12}\)=9,1
b) x = \(\dfrac{2,5.12}{1.5}\) = 20
c) Nhân chéo: 7.(x-1) = 6.(x+5)
<=> 7x - 7 = 6x +30
<=> 7x - 6x = 7 + 30 (chuyển vế)
-> x = 37
d) Nhân chéo: 25x2 = 24.6 = 144
x2 = \(\dfrac{144}{25}\)=5,76
-> x = \(\sqrt{5,76}\) = 2,4
e) Nhân chéo: (x-2)2 = 4.9 = 36
Ta dễ thấy (x-2)2 = 62
-> x-2 = 6 -> x = 6+2 = 8
TICK NHÉ :)
\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)
\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)
\(a,=\dfrac{9}{12}-\dfrac{7}{12}-\dfrac{8}{12}=-\dfrac{1}{2}\\ b,=\dfrac{95}{6}+\dfrac{17}{18}-\dfrac{5}{12}=\dfrac{589}{36}\)
\(a,=\dfrac{1}{6}-\dfrac{5}{6}=-\dfrac{4}{6}=-\dfrac{2}{3}\\ b,=\dfrac{3^2}{3^{12}\cdot3^{18}}=\dfrac{1}{3^{28}}\)
a) Ta xét tỉ số \(\dfrac{6}{5}:2 = \dfrac{6}{5}.\dfrac{1}{2} = \dfrac{6}{{10}} = \dfrac{3}{5}\)
Tương tự xét với tỉ số \(\dfrac{{12}}{5}:4 = \dfrac{{12}}{5}.\dfrac{1}{4} = \dfrac{{12}}{{20}} = \dfrac{{12:4}}{{20:4}} = \dfrac{3}{5}\)
Ta thấy các tỉ số đều bằng \(\dfrac{3}{5}\) nên ta sẽ lập được một tỉ lệ thức : \(\dfrac{{12}}{5}:4\) = \(\dfrac{6}{5}:2\)
b) Từ các số 9;2;3;6 ta thấy :
\(\dfrac{9}{3}\)= 3 và \(\dfrac{6}{2}\)=3 nên suy ra ta có tỉ lệ thức thứ nhất : \(\dfrac{9}{3}\)=\(\dfrac{6}{2}\)
Ta xét tỉ số \(\dfrac{9}{6}\)=\(\dfrac{{9:3}}{{6:3}}\)=\(\dfrac{3}{2}\)nên ta có được tỉ lệ thức thứ hai : \(\dfrac{9}{6}\)=\(\dfrac{3}{2}\)
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) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
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:
\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)
\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)
\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
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