\(a.2.x-138=2^3.3^2\)

    \(2.x-138=8.9\)

    \(2.x-138=72\)

    \(2.x\)          \(=72+138\)

    \(2.x\)          \(=210\)  

       \(x\)           \(=210:2\)  

       \(x\)            \(=105\)

\(b.231-\left(x-6\right)=1339:13\)

    \(231-\left(x-6\right)=103\)

                    \(x-6=231-103\)

                    \(x-6=128\)

                    \(x\)      \(=128+6\) 

                    \(x\)       \(=134\)

\(c.\left[\left(6.x-72\right):2-84\right].28=5628\)

    \(\left[\left(6.x-72\right):2-84\right]\)   \(=5628:28\)

       \(\left(6.x-72\right):2-84\)      \(=201\)

          \(\left(6.x-72\right):2\)               \(=201-84\)

           \(\left(6.x-72\right):2\)              \(=117\)

              \(6.x-72\)                      \(=117.2\)

               \(6.x-72\)                      \(=234\)

               \(6.x\)                                  \(=234+72\)   

               \(6.x\)                                   \(=306\)    

                   \(x\)                                     \(=306:6\)

                  \(x\)                                       \(=51\)

mik nè

Cam on nha

`#3107.101107`

a,

\(\text{A = }\left\{x\in R\text{ | }\left(2x-x^2\right)\left(3x-2\right)=0\right\}\)

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

`<=>`\(\left[{}\begin{matrix}2x-x^2=0\\3x-2=0\end{matrix}\right.\)

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

`<=>`\(\left[{}\begin{matrix}x=0\\2-x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy, `A = {0; 2; 2/3}`

b,

\(\text{B = }\left\{x\in R\text{ | }2x^3-3x^2-5x=0\right\}\)

`<=> 2x^3 - 3x^2 - 5x = 0`

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

`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-3x-5=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-2x+5x-5=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x^2-2x\right)+\left(5x-5\right)=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2x\left(x-1\right)+5\left(x-1\right)=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2x+5=0\\x-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

Vậy, `B = {-5/2; 0; 1}.`

c,

\(\text{C = }\left\{x\in Z\text{ | }2x^2-75x-77=0\right\}\)

`<=> 2x^2 - 75x - 77 = 0`

`<=> 2x^2 - 2x + 77x - 77 = 0`

`<=> (2x^2 - 2x) + (77x - 77) = 0`

`<=> 2x(x - 1) + 77(x - 1) = 0`

`<=> (2x + 77)(x - 1) = 0`

`<=>`\(\left[{}\begin{matrix}2x+77=0\\x-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}2x=-77\\x=1\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=-\dfrac{77}{2}\\x=1\end{matrix}\right.\)

Vậy, `C = {-77/2; 1}`

d,

\(\text{D = }\left\{x\in R\text{ | }\left(x^2-x-2\right)\left(x^2-9\right)=0\right\}\)

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

`<=>`\(\left[{}\begin{matrix}x^2-x-2=0\\x^2-9=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2+x-2x-2=0\\x^2=9\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}\left(x^2+x\right)-\left(2x+2\right)=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x\left(x+1\right)-2\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x=\pm3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=2\\x=-1\\x=\pm3\end{matrix}\right.\)

Vậy, `D = {-1; -3; 2; 3}.`

Tương tự câu 1. HS tự làm

Tu ten suy ra x thuoc UC(84;180)

84=2^2x3x7

180=2^2x3^2x5

=>UCLN(84;180)=2^2x3=12

=>UC(84;180)=U(12)={1;2;3;4;6;12}

Vi x>6 nen suy ra x=12