a, \(\left(x-1\right).\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

b, \(\left(2x-4\right).\left(3x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\3x+9=0\end{matrix}\right.\left[{}\begin{matrix}2x=4\\3x=-9\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a) TH₁: x-1=0 => x=1

     TH₂: x+2=0 => x=-2

b) TH₁: 2x-4=0 <=> 2x= 4 <=> x=2

     TH₂: 3x+9=0 <=> 3x=-9 <=> x= -3

a) ( x - 1) . ( x + 2 ) = 0 

=> x - 1 = 0 => x = 1

     x + 2 = 0     x = -2 

vậy x ϵ { 1; -2 }

b) ( 2x - 4 ) . ( 3x + 9 ) = 0

=> 2x - 4  = 0   => 2x = 4  => x = 2

     3x + 9 = 0        3x = -9      x = -3

vậy x ϵ { 2 ; -3 {

P/S : phần mình suy ra thì bn đóng ngoặc vuông to rồi mới ghi phép tính nhé!

5            suy ra n+1chia hết n-5

              suy ra (n+1)-(n-5)chia hết n-5

              tương đương n+1-n+5 chia hết n-5

             tương đương 6 chia hết n-5

            suy ra n-5 thuộc vào Ư6=1,2,3,6,-1,-2,-3,-6

            suy ra n thuộc vào =6,7,8,11,4,3,2,-1

Trl

-Bạn kia  làm đúng r nhé !~ :>

Học tốt 

nhé bạn ~

học tốt nhé

a, 7\(x\).(\(x\) - 10) = 0

   \(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

Vậy \(x\in\) {0; 10}

b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0

    \(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)

       \(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)

c, (4 - 2\(x\)).(\(x\) - 3) = 0

    \(\left[{}\begin{matrix}4-2x=0\\x-3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}2x=4\\x=3\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4:2\\x=3\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

    Vậy \(x\in\) {2; 3}

\(\left(2x-1\right).\left(2x-5\right)< 0.\)

Vì \(2x-1>2x-5\)nên

\(\Rightarrow\hept{\begin{cases}2x-1>0\\2x-5< 0\end{cases}\Rightarrow\hept{\begin{cases}2x>1\\2x< 5\end{cases}\Rightarrow}}\hept{\begin{cases}x>\frac{1}{2}\\x< \frac{5}{2}\end{cases}\Leftrightarrow\frac{1}{2}< x< \frac{5}{2}}\)

Vậy \(\frac{1}{2}< x< \frac{5}{2}\)thỏa mãn đề bài

a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

b. \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)

c, \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)

\(\Rightarrow5x=7\)

\(\Rightarrow x=\frac{7}{5}\)

e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }

 Vậy....

a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

Vậy : ....

b) \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)

c) \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

Vậy :...

`\color{grey}\text{#071931}`

`(4 + 2x) * (9 - 3x) = 0`

TH1: `4 + 2x = 0 => 2x = -4 => x = -2`

TH2: `9 - 3x = 0 => 3x = 9 => x = 3`

Vậy, `x \in {-2; 3}`

____

`(3x - 19)^3 = 125`

`=> (3x - 19)^3 = 5^3`

`=> 3x - 19 = 5`

`=> 3x = 24`

`=> x = 8`

Vậy, `x = 8.`