a) Để \(\frac{13}{x-1}\)là số nguyên thì 13 chia hết cho x-1

=> x-1 thuộc Ư (13)={-13;-1;1;13}
Ta có bảng

b) Có x+3=x-2+5

=> 5 chia hết cho x-2

=> x-2 thuộc Ư(5)={-5;-1;1;5}

Ta có bảng

c) 2x=2(x+2)-4

=> 4 chia hết cho x+2 hay x+2 thuộc Ư (4)={-4;-2;-1;1;2;4}

Ta có bảng

\(\left(2-\frac{x}{13}\right)^3=\frac{1331}{2197}\)

=> \(\left(2-\frac{x}{13}\right)^3=\left(\frac{11}{13}\right)^3\)

=> \(2-\frac{x}{13}=\frac{11}{13}\)

=> \(\frac{x}{13}=2-\frac{11}{13}=\frac{2\cdot13-11}{13}=\frac{15}{13}\)

=> x = 15

\(X:\dfrac{3}{9}=\dfrac{7}{8}\)

\(X=\dfrac{7}{8}\times\dfrac{3}{9}\)

\(X=\dfrac{21}{72}\)

\(\left(X-21\times13\right):11=30\)

\(X-273=330\)

\(X=603\)

`x:3/9=7/8`

`x=7/8xx3/9`

`x=7/24`

`-------------`

`(x-21xx13):11=30`

`(x-273):11=30`

`x-273=30xx11`

`x-273=330`

`x=330-273`

`x=57`

\(a,x-1⋮x-3\)

\(\Rightarrow x-3+2⋮x-3\)

\(\Rightarrow2⋮x-3\)

\(x-3=\left\{-2;-1;1;2\right\}\)

\(x=\left\{1;2;4;5\right\}\)

\(b,x+6⋮x-1\)

\(\Rightarrow x-1+7⋮x-1\)

\(\Rightarrow7⋮x-1\)

\(x=\left\{-6;0;2;8\right\}\)

\(c,x⋮x-5\)

\(x-5+5⋮x-5\)

\(5⋮x-5\)

\(x=\left\{0;4;6;11\right\}\)

\(x^5+x^4+x+1=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

bạn giải thích ngắn gọn quá

x:x*15-3,7=2,3

x:2*15-3,7=2,3

x:2*15=2,3+3,7

x:2*15=6

x:2=6:15

x*2=0,4

x=0,4:2

x=0,2

Bài 1:

a: \(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)

=>\(\left(\frac{x^2+x+4}{2}-3\right)+\left(\frac{x^2+x+7}{3}-3\right)=\left(\frac{x^2+x+13}{5}-3\right)+\left(\frac{x^2+x+16}{6}-3\right)\)

=>\(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)

=>\(\left(x^2+x-2\right)\left(\frac12+\frac13-\frac15-\frac16\right)=0\)

=>\(x^2+x-2=0\)

=>(x+2)(x-1)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=1\end{array}\right.\)

b: \(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{25}\)

=>\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{25}\)

=>\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{25}\)

=>\(\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{25}\)

=>\(\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{25}\)

=>\(\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{25}\)

=>(x+1)(x+10)=25

=>\(x^2+11x+10-25=0\)

=>\(x^2+11x-14=0\)

=>\(x^2+11x+\frac{121}{4}=44,25\)

=>\(\left(x+\frac{11}{2}\right)^2=\frac{177}{4}\)

=>\(x+\frac{11}{2}=\pm\sqrt{\frac{177}{2}}\)

=>\(x=-\frac{11}{2}\pm\frac{\sqrt{177}}{2}\)

x(x-3)+(x-3)=0

=>(x-3)(x+1)=0

=>\(\left[\begin{array}{l}x-3=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-1\end{array}\right.\)