Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$

$x-x\times \frac{1}{2}=\frac{3}{2}$

$x\times (1-\frac{1}{2})=\frac{3}{2}$

$x\times \frac{1}{2}=\frac{3}{2}$

$x=\frac{3}{2}: \frac{1}{2}=3$

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

B=2x^2+6x-3x^2-3=x+1-x(x-2)=0

  =-x^2+6x-3=x+1-x^2+x=0

  =4x-3=0

x=3/4

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)

Suy ra: \(x^2-4x+3-x^2=0\)

\(\Leftrightarrow-4x=-3\)

hay \(x=\dfrac{3}{4}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{3}{4}\right\}\)

\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)

\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(=x^3+3x^2+3x+1-x^3+1-2=0\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow3x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

câu a ) a*x^19+1 

câu b ) 

đa thức chia có bậc 2 nên đa thức dư có bậc không quá 1. vậy đa thức dư có bậc nhất dạng ax+b

Ta có: x67+x47+x27+x7+x+1=(x2−1).Q(x)+ax+bx67+x47+x27+x7+x+1=(x2−1).Q(x)+ax+b

Cho x=1 rồi x=-1 ta được: \hept{1+1+1+1+1+1=a+b−1−1−1−1−1+1=−a+b\hept{1+1+1+1+1+1=a+b−1−1−1−1−1+1=−a+b

⇔\hept{a+b=6−a+b=−4⇔\hept{a=5b=1⇔\hept{a+b=6−a+b=−4⇔\hept{a=5b=1

Vậy dư trong phép chia trên là 5x+1

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

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

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

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

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

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

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

\(\Rightarrow x=\dfrac{-13}{12}\)

\(\left(\dfrac{2x}{3}-\dfrac{1}{3}\right)+\left(3x-2x+1\right)=8\)

\(\Leftrightarrow\dfrac{2x-1}{3}+x-7=0\Rightarrow2x-1+3x-21=0\Leftrightarrow x=\dfrac{22}{5}\)

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left[3x-2\left(x-1\right)\right]=8\)

\(\Rightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+2=8\)

\(\Rightarrow\dfrac{5}{3}x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{5}\)

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left(3x-2\left(x-1\right)\right)=8\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+1=8\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{22}{3}\)

\(\Leftrightarrow x=\dfrac{22}{5}\)

a: ĐKXĐ: x<>0; x<>-1

PT =>x+1-2x=3

=>1-x=3

=>x=-2(nhận)

b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

=>x-3=5(2x-3)

=>10x-15=x-3

=>9x=12

=>x=4/3(nhận)

c: ĐKXĐ: x<>0; x<>2

PT =>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=-1