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

\(\Leftrightarrow x+1=0\)

hay x=-1

2x - 1 + 4x - 2 + 6x - 3 + ... + 400x - 200 = 5 + 10 + 15 + ... + 10000

=> (2x + 4x + 6x + ... + 400x) - (1 + 2 + 3 + ... + 200) = 5.(1 + 2 + 3 + ... + 200)

=> 2x.(1 + 2 + 3 + ... + 200) - (1 + 2 + 3 + ... + 200) = 5.(1 + 2 + 3 + ... + 200)

=> (1 + 2 + 3 + ... + 200).(2x - 1) = 5.(1 + 2 + 3 + ... + 200)

=> 2x - 1 = 5

=> 2x = 5 + 1

=> 2x = 6

=> x = 6 : 2

=> x = 3

Vậy x = 3

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

\(a,x-1⋮x-3\)

\(\Rightarrow x-3+2⋮x-3\)

\(\Rightarrow2⋮x-3\)

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

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

\(b,x+6⋮x-1\)

\(\Rightarrow x-1+7⋮x-1\)

\(\Rightarrow7⋮x-1\)

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

\(c,x⋮x-5\)

\(x-5+5⋮x-5\)

\(5⋮x-5\)

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

\(x^3-5x^2-9x+45=0\)

=>\(x^2\left(x-5\right)-9\left(x-5\right)=0\)

=>\(\left(x-5\right)\left(x^2-9\right)=0\)

=>\(\left(x-5\right)\left(x-3\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\\x=-3\end{matrix}\right.\)

`x^3 -5x^2 -9x+45=0`

`<=> (x^3 -5x^2 )-(9x-45)=0`

`<=> x^2 (x-5)- 9(x-5)=0`

`<=>(x-5)(x^2 -9)=0`

`<=>(x-5)(x-3)(x+3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-3=0\\x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\\x=-3\end{matrix}\right.\)

x = 4/5 : 3/4

x = 16/15

x = 5/9 : 8/3

x = 5/9 x 3/8

x = 5/24

\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)

\(x=\dfrac{16}{15}\)

-----------------------

\(x=\dfrac{5}{9}\times\dfrac{3}{8}\)

\(x=\dfrac{5}{24}\)

\(x\times\dfrac{3}{4}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}:\dfrac{3}{4}\)

\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)

\(x=\dfrac{16}{15}\)

\(\dfrac{5}{9}:x=\dfrac{8}{3}\)

      \(x=\dfrac{5}{9}:\dfrac{8}{3}\)

      \(x=\dfrac{5}{9}\times\dfrac{3}{8}\)

     \(x=\dfrac{15}{72}\)

    \(x=\dfrac{5}{24}\)

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