\(a,\dfrac{2}{3}.x=\dfrac{2}{7}\\ x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{3}{7}\\ ---\\ b,x.\dfrac{3}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\\ ---\\ c,x:\dfrac{8}{13}=\dfrac{13}{7}\\x=\dfrac{13}{7}.\dfrac{8}{13}=\dfrac{8}{7}\\ ----\\ d,\dfrac{3}{2}:x=\dfrac{7}{4}\\ x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{3}{2}.\dfrac{4}{7}=\dfrac{6}{7}\)

\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)

\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)

\(\rightarrow10x+80+15x+105=-6x\)

\(\Leftrightarrow31x+185=0\)

\(\Leftrightarrow x=-\frac{185}{31}\)

b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)

\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)

\(\rightarrow20x-160+15x-105=240+12-12x\)

\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)

2/5 + x = 15/20

2/5 + x = 3/4

x = 3/4 - 2/5 

x = 7/20

8/9 - 2/9 : x = 2/3

2/9 : x = 8/9 - 2/3 

2/9 : x = 2/9

x = 2/9 : 2/9

x = 1

\(\frac{2}{5}+x=\frac{15}{20}\)

\(\frac{2}{5}+x=\frac{3}{4}\)

\(x=\frac{3}{4}-\frac{2}{5}\)

\(x=\frac{7}{20}\)

\(\frac{8}{9}-\frac{2}{9}:x=\frac{2}{3}\)

\(\frac{2}{9}:x=\frac{8}{9}-\frac{2}{3}\)

\(\frac{2}{9}:x=\frac{2}{9}\)

\(x=\frac{2}{9}:\frac{2}{9}\)

\(x=1\)

Mik làm chi tiết nhé, bạn có thể lược 1 số phần nếu thích

\(\frac{2}{5}+x=\frac{15}{20}\)

=>\(\frac{2}{5}+x=\frac{3}{5}\)

=>\(x=\frac{3}{5}-\frac{2}{5}\)

=> \(x=\frac{1}{5}\)

\(\frac{8}{9}-\frac{2}{9}:x=\frac{2}{3}\)

=> \(\frac{8}{9}-\frac{2}{9}:x=\frac{6}{9}\)

=> \(\frac{2}{9}:x=\frac{8}{9}-\frac{6}{9}\)

=>\(\frac{2}{9}:x=\frac{2}{9}\)

=> \(x=\frac{2}{9}:\frac{2}{9}\)

=> \(x=1\)

A. x = 2

B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)

C. x = 3

D. \(x=\dfrac{4.6}{8}=3\)

E. \(x=\dfrac{7}{3}\)

G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)

<=>\(14\left(10-x\right)=364\)

<=> 10 - x = 26 

<=> x = -16 

H. \(3\left(x+2\right)=4\left(x-5\right)\)

<=> 3x + 6  = 4x - 20 

<=> -x = -26

<=> x = 26

K. \(\dfrac{x}{2}=\dfrac{8}{x}\)

<=> \(x^2=16\)

<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

M. \(\left(x-2\right)^2=100\)

<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a=2

b=16

c=3

d=3

mik chỉ biết thế này thôi(ko chắc đúng=3)

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

B,\(\dfrac{3}{8}=\dfrac{6}{x};\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16}\Rightarrow x=16\)

C,\(\dfrac{1}{9}=\dfrac{x}{27};\dfrac{1}{9}=\dfrac{1\cdot3}{9\cdot3}=\dfrac{3}{27}\Rightarrow x=3\)

D,\(\dfrac{4}{x}=\dfrac{8}{6};\dfrac{4}{x}=\dfrac{4\cdot2}{x\cdot2}=\dfrac{8}{2x}\)

   \(\Rightarrow2x=6;6=x:2;x=3\)

,làm ơn giúp mik với ah

\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

= \(\dfrac{337.2023}{2}\)

= \(\dfrac{\text{681751}}{2}\)

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

⇔\(\left(x+4\right)\left(x+4\right)=100\)

⇔\(\left(x+4\right)^2=10^2\)

⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

a: \(-\dfrac{2}{5}+\dfrac{4}{5}x=\dfrac{3}{5}\)

=>\(\dfrac{4}{5}x=\dfrac{3}{5}+\dfrac{2}{5}=1\)

=>\(x=1:\dfrac{4}{5}=\dfrac{5}{4}\)

b; \(-\dfrac{3}{7}-\dfrac{4}{7}:x=-2\)

=>\(\dfrac{4}{7}:x+\dfrac{3}{7}=2\)

=>\(\dfrac{4}{7}:x=2-\dfrac{3}{7}=\dfrac{11}{7}\)

=>\(x=\dfrac{4}{7}:\dfrac{11}{7}=\dfrac{4}{11}\)

Ta có: \(\left(3x+2\right)^3-3x\left(3x+4\right)^2-17x\left(x-3\right)=-54\)

\(\Leftrightarrow27x^3+54x^2+36x+8-3x\left(9x^2+24x+16\right)-17x^2+51x=-54\)

\(\Leftrightarrow27x^3+37x^2+87x+8+54-27x^3-72x^2-48x=0\)

\(\Leftrightarrow-35x^2+39x+62=0\)

\(\Delta=39^2-4\cdot\left(-35\right)\cdot62=10201\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-39-101}{-70}=\dfrac{-140}{-70}=2\\x_2=\dfrac{-39+101}{-70}=\dfrac{-62}{70}=\dfrac{-31}{35}\end{matrix}\right.\)