Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-2}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-2}\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=100\)

\(\Leftrightarrow x^2-5x+6-100=0\)

\(\Leftrightarrow x^2-5x-94=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{401}}{2}\\x=\dfrac{5+\sqrt{401}}{2}\end{matrix}\right.\)

\(3x^2-5x+3=2\)

\(5\left(3x^2-5x+3\right)=2.5\)

=>\(15x^2-25x+15=10\)

\(15x^2-25x+15+3=10+3\)

\(15x^2-25x+18=13\)

15x²-25x+18

=(3x².5-5.5x+15)+3

=5(3x²-5x+3)+3

=5.2+3

=10+3=13

spam nha, lm cho r mà

\(a,\Rightarrow n-2+5⋮n-2\\ \Rightarrow n-2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-3;1;3;7\right\}\\ b,\Rightarrow2\left(n-4\right)+13⋮n-4\\ \Rightarrow n-4\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\\ \Rightarrow n\in\left\{-9;3;5;17\right\}\\ c,\Rightarrow6n-9⋮3n+1\\ \Rightarrow2\left(3n+1\right)-12⋮3n+1\\ \Rightarrow3n+1\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\\ \Rightarrow n\in\left\{-1;0;1\right\}\left(n\in Z\right)\\ d,\Rightarrow n^2+2n-n-2+3⋮n+2\\ \Rightarrow n\left(n+2\right)-\left(n+2\right)+3⋮n+2\\ \Rightarrow n+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow n\in\left\{-5;-3;-1;1\right\}\)

\(x-\dfrac{3}{5}=\dfrac{2}{3}\times\dfrac{1}{5}\\ \Rightarrow x-\dfrac{3}{5}=\dfrac{2}{15}\\ \Rightarrow x=\dfrac{2}{15}+\dfrac{3}{5}\\ \Rightarrow x=\dfrac{2}{15}+\dfrac{9}{15}\\ \Rightarrow x=\dfrac{11}{15}\)

=>x-3/5=2/15

hay x=11/15

x-3/5=2/3 x1/5

x-3/5=2/15

x=2/15+3/5

x=2/15+9/15

x=11/15

x=599

x = vo nghiem

ứ hoàng anh nói đúng rồi đó!

có cái đầu buồi

\(\frac{1}{3}\)- l 5 - x l = \(\frac{2}{3}\)

=> l 5 - x l = \(\frac{-1}{3}\)

mà l 5 - x l \(\ge\)0 với mọi x thuộc Z

=> không có giá trị x thỏa mãn đề bài

có số thoả mãn x mà bạn

x=4

à mk nhầm

1/3-|5-x|=2-2/3

sorry

\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)

\(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6z^2=5\)

\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

Vậy x = 1