a, \(2,5:7,5=x:\dfrac{3}{5}\)

\(\Leftrightarrow x:\dfrac{3}{5}=2,5:7,5\)

=> \(x.7,5=\dfrac{3}{5}.2,5\) => \(x=\dfrac{1,5}{7,5}=\dfrac{1}{5}\)

b, \(2\dfrac{2}{3}:x=1\dfrac{7}{9}\)

=> \(x=2\dfrac{2}{3}:1\dfrac{7}{9}\)

=> \(x=\dfrac{3}{2}\)

c, \(\dfrac{5}{6}:x=20:3\)

=> \(x.20=\dfrac{5}{6}.3\) => \(x=\dfrac{2,5}{20}=\dfrac{1}{8}\)

3, Tìm x, biết:

a) 2,5 : 7,5 = x : \(\dfrac{3}{5}\)

=> \(x:\dfrac{3}{5}=\dfrac{1}{3}\)

=> \(x=\dfrac{1}{3}.\dfrac{3}{5}=>x=\dfrac{1}{5}\)

b) \(2\dfrac{2}{3}\) : x = \(1\dfrac{7}{9}\)

=> \(\dfrac{8}{3}:x=\dfrac{16}{9}\)

=> \(x=\dfrac{8}{3}:\dfrac{16}{9}=>x=\dfrac{3}{2}\)

c) \(\dfrac{5}{6}\) : x = 20 : 3

=> \(x=\dfrac{5}{6}:\dfrac{20}{3}=>x=\dfrac{1}{8}\)

a) \(\dfrac{27}{2}\cdot7,5+\dfrac{27}{2}\cdot2,5-150\)

\(=\dfrac{27}{2}\cdot\left(7,5+2,5\right)-150\)

\(=\dfrac{27}{2}\cdot10-150\)

\(=135-150\)

\(=-15\)

b) \(3^3\cdot\dfrac{18}{5}-3^3\cdot2\dfrac{2}{5}-3^3\cdot\dfrac{6}{5}\)

\(=3^3\cdot\dfrac{18}{5}-3^3\cdot\dfrac{12}{5}-3^3\cdot\dfrac{6}{5}\)

\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{12}{5}-\dfrac{6}{5}\right)\)

\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{18}{5}\right)\)

\(=3^3\cdot0\)

\(=0\)

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10

a) \(2,5:7,5=x:\dfrac{3}{5}\)

\(\Rightarrow\dfrac{2,5}{7,5}=\dfrac{x}{\dfrac{3}{5}}\)

\(\Rightarrow x=\dfrac{\dfrac{3}{5}\cdot2.5}{7,5}\)

\(\Rightarrow x=\dfrac{1}{5}\) 

b) \(\dfrac{5}{6}:x=20:3\)

\(\Rightarrow\dfrac{\dfrac{5}{6}}{x}=\dfrac{20}{3}\)

\(\Rightarrow x=\dfrac{3\cdot\dfrac{5}{6}}{20}\)

\(\Rightarrow x=\dfrac{1}{8}\)

c) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}\)

\(\Rightarrow\dfrac{8}{3}:x=\dfrac{16}{9}\)

\(\Rightarrow\dfrac{\dfrac{8}{3}}{x}=\dfrac{16}{9}\)

\(\Rightarrow x=\dfrac{\dfrac{8}{3}\cdot9}{16}\)

\(\Rightarrow x=\dfrac{3}{2}\)

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)

=> 7x+ 35 = 3x - 18

7x - 3x = -18 -35

4x = -53

x = -53:4

x = \(\dfrac{-53}{4}\)

\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)

\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)