Đầu bài phải là tìm x thuộc Z sao cho phân số thuộcZ chứ

a)Để phân số \(\frac{2x+1}{x-3}\in Z\)

 \(\Leftrightarrow2x+1⋮x-3\)

\(\Leftrightarrow2x-6+7⋮x-3\)

\(\Leftrightarrow2\left(x-3\right)+7⋮x-3\)

mà \(2\left(x-3\right)⋮x-3\)

\(\Rightarrow7⋮x-3\)

\(\Rightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Tìm nốt

thank you

a)5-2x=3x+20

   5=3x+20+2x

   5=5x+20

=>5x+20=5

    5x=5-20

    5x=-15

   x=(-15):5

  x=-3

Bùi Ngọc Truongf Sơn làm nốt cho mình 2 bài còn lại đi

\(a,5-2x=3x+20\)

\(\Leftrightarrow-2x-3x=20-5\)

\(\Leftrightarrow-5x=15\)

\(\Leftrightarrow x=-3\)

\(b,5-\left(1-3x\right)=2x+4\)

\(\Leftrightarrow5-1+3x=2x+4\)

\(\Leftrightarrow5-1-4=2x-3x\)

\(\Leftrightarrow0=-x\)

\(\Leftrightarrow x=0\)

\(c,35-\left(5-2x\right)=7-\left(-3+3x\right)\)

\(\Leftrightarrow35-5+2x=7+3-3x\)

\(\Leftrightarrow35-5-7-3=-2x-2x\)

\(\Leftrightarrow20=-4x\)

\(\Leftrightarrow x=-5\)

\(\dfrac{2\text{x}-1}{3}=\dfrac{3\text{x}+1}{4}\)

\(\Leftrightarrow=\dfrac{4\left(2\text{x}-1\right)}{12}=\dfrac{3\left(3\text{x}+1\right)}{12}\)

\(\Leftrightarrow8\text{x}-4=9\text{x}+3\)

\(\Leftrightarrow8\text{x}-9\text{x}=3+4\)

\(\Leftrightarrow-x=7\)

\(\Leftrightarrow x=-7\)

`(2x-1)/3 = (3x+1)/4`

`=> (2x-1).4= 3.(3x+1)`

`=> 8x -4= 9x+3`

`=> 8x-9x =3+4`

`=> -x=7`

`=>x=-7`

\(2\left(x-2\right)\left(\sqrt[3]{4x-4}+\sqrt{2x-2}\right)=3x-1\)

\(\Leftrightarrow2\left(x-2\right)\left[\left(\sqrt[3]{4x-4}-2\right)+\left(\sqrt{2x-2}-2\right)\right]+8\left(x-2\right)=3x-1\)

\(\Leftrightarrow2\left(x-2\right)\left[\frac{4x-12}{\sqrt[3]{\left(4x-4\right)^2}+2\sqrt[3]{4x-4}+4}+\frac{2x-6}{\sqrt{2x-2}+2}\right]+\left(5x-15=0\right)\)

\(\left(x-3\right)\left[\frac{8\left(x-2\right)}{...}+\frac{4\left(x-2\right)}{...}+5\right]=0\Leftrightarrow x=3.\)

1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)

Vậy ...................

b/ ĐKXĐ:\(x\ne2;x\ne5\)

.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x^2-10x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)

Vậy ..............

`Answer:`

`1.`

a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)

b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)

\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)

`2.`

\(ĐKXĐ:x\ne-m-2;x\ne m-2\)

Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)

a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)

b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì

\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)

\(3^{x+4}=9^{2x-1}\)

\(\Rightarrow3^{x+4}=3^{4x-2}\)

\(\Rightarrow x+4=4x-2\)

\(\Rightarrow3x=6\Rightarrow x=2\)

3x+4=92x−13x+4=92x−1

⇒3x+4=34x−2⇒3x+4=34x−2