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

Cảm ơn ạ

1. áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x+2}{3}=\frac{y-7}{5}=\frac{x+y-5}{3+5}=\frac{16}{8}=2\Rightarrow\hept{\begin{cases}x+2=6\\y-7=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=17\end{cases}}}\)

2. áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-y+2}{2-3}=\frac{-10+7}{-1}=3\Rightarrow\hept{\begin{cases}x+5=6\\y-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=11\end{cases}}\)

`5(x+2) -2x=18`

`=> 5x+10 -2x=18`

`=> 3x+10=18`

`=>3x=18-10`

`=>3x=8`

`=>x=8/3`

=>5x+10-2x=18

=>3x+10=18

=>3x=18-10=8

=>x=\(\dfrac{8}{3}\)

\(\Leftrightarrow x:\dfrac{2}{3}=\dfrac{17}{5}:\dfrac{5}{2}=\dfrac{17}{5}\cdot\dfrac{2}{5}=\dfrac{34}{25}\)

hay \(x=\dfrac{68}{75}\)

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

\(\Rightarrow3\left(x+2\right)^2=2.5^2+5^2\)

\(\Rightarrow3\left(x+2\right)^2=5^2\left(2+1\right)\)

\(\Rightarrow3\left(x+2\right)^2=5^2.3\)

\(\Rightarrow\left(x+2\right)^2=5^2\)

\(\Rightarrow\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\) \(\Rightarrow x=3\left(x\inℕ\right)\)

3(x + 2)² - 5² = 2.5²

3(x + 2)² - 25 = 50

3(x + 2)² = 50 + 25

3(x + 2)² = 75

(x + 2)² = 75 : 3

(x + 2)² = 25

x + 2 = 5 hoặc x + 2 = -5

*) x + 2 = 5

x = 5 - 2

x = 3 (nhận)

*) x + 2 = -5

x = -5 - 2

x = -7 (loại)

Vậy x = 3

pt vt lại:

\(\dfrac{x+4}{x^2-3x+2}+\dfrac{x+1}{x^2-4}+5=\dfrac{2x+5}{x^2-4x+5}\)

pt này đk?

\(\left(x+\frac{2}{5}\right)^5=\left(x+\frac{2}{5}\right)^3\)

\(\Leftrightarrow\left(x+\frac{2}{5}\right)^3\left[\left(x+\frac{2}{5}\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{5}=0\\x+\frac{2}{5}=\pm1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{7}{5};x=-\frac{3}{5}\end{cases}}\)

a: \(3x^2\left(x^2-2x+5\right)\)

\(=3x^2\cdot x^2-3x^2\cdot2x+3x^2\cdot5\)

\(=3x^4-6x^3+15x^2\)

b: (x+1)(2x-3)

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

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

c: \(\left(2x^3-3x^2+x+15\right):\left(2x+3\right)\)

\(=\left(2x^3+3x^2-6x^2-9x+10x+15\right):\left(2x+3\right)\)

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

16:Sửa đề: \(\frac12x+\frac16\left(x-2\right)=\frac34-2x\)

=>\(\frac12x+\frac16x-\frac26=\frac34-2x\)

=>\(\frac46x-\frac13=\frac34-2x\)

=>\(\frac23x+2x=\frac34+\frac13\)

=>\(\frac83x=\frac{9}{12}+\frac{4}{12}=\frac{13}{12}\)

=>\(x=\frac{13}{12}:\frac83=\frac{13}{12}\times\frac38=\frac{13}{4\times8}=\frac{13}{32}\)

19: \(\frac{5}{12}x+3=\frac13-\frac{7}{12}x\)

=>\(\frac{5}{12}x+\frac{7}{12}x=\frac13-3\)

=>\(\frac{12}{12}x=\frac13-\frac93=-\frac83\)

=>\(x=-\frac83\)

20: \(\frac12x+\frac52=\frac72x-\frac34\)

=>\(\frac12x-\frac72x=-\frac34-\frac52\)

=>\(-3x=-\frac34-\frac{10}{4}=-\frac{13}{4}\)

=>\(3x=\frac{13}{4}\)

=>\(x=\frac{13}{4}:3=\frac{13}{12}\)