\(\Leftrightarrow x\in\left\{0;43\right\}\)

a)  \(645-\left(27+x\right)-342=123\)

\(\Leftrightarrow\)\(-24-x=123\)

\(\Leftrightarrow\)\(x=123+24=147\)

Vậy...

b)  \(\left|x\right|+87=123\)

\(\Leftrightarrow\)\(\left|x\right|=123-87=36\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=36\\x=-36\end{cases}}\)

Vậy....

c)  \(\left|x-12\right|=42\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-12=42\\x-12=-42\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=54\\x=-30\end{cases}}\)

Vậy...

a) 645 - ( 27 + x ) - 342 = 123

    645 - ( 27 + x )           = 123 + 342

    645 - ( 27 + x )           =  465

               27 + x              = 645 - 465

               27 + x              =   180

                       x              = 180 - 27

                       x              =  53

b) ! x ! + 87 = 123

    ! x !          = 123 - 87

    ! x !          =  36

 -> x = 36 hoặc x = -36

c) ! x - 12 ! = 42

-> x - 12 = 42 hoặc x - 12 = -42

Nếu x - 12 = 42 thì ta có :                                     Nếu x - 12 = -42 thì ta có :

        x  = 42 + 12                                                          x =  -42 + 12

        x =      54                                                              x =       -30

a, => 645-27-x-342=123

=> 276-x=123

=> x=276-123 = 153

a, (-72).(15 - 49) + 15.(-56 + 72)

= -72 . 15 + 72.49  - 15.56 + 15.72

= (-72.15 + 15.72) + (72.49  - 15.56)

= 3528 - 84

= 2688

1b, 1532 + (-186) + (-1432) + (-14) + 123

= (1532 - 1432) - (186 + 14) + 123

= 100 - 200 + 123

= 223 - 200

= 23 

43.(53 - 81) + 53.(81 - 43)

= 43.53 - 81.43 + 53.81 - 53.43

= (43.53 - 53.43) + (81.53 - 81.43)

= 0 + 81.(53-43)

= 81.10

= 810

a. 312; 318; 324; 330; 336.

b. 165; 180; 195; 210.

\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)

\(\Leftrightarrow\dfrac{x+35}{65}+1+\dfrac{x+39}{61}+1=\dfrac{x+43}{57}+1+\dfrac{x+47}{53}+1\)

\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\right)=0\Leftrightarrow x=-100\)

Ta có:

\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)

Ta thấy:

\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)

Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)

\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)

 Vậy \(x=-100\)

Ta có: \(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)

\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)

\(\Leftrightarrow x+100=0\)

hay x=-100

\(x\left(x+1\right)=156\)

\(\Rightarrow x^2+x=156\)

\(\Rightarrow x^2+x-156=0\)

\(\Rightarrow x^2+13x-12x-156=0\)

\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)

\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)

___________________

\(x\left(x+1\right)=342\)

\(\Rightarrow x^2+x=342\)

\(\Rightarrow x^2+x-342=0\)

\(\Rightarrow x^2+19x-18x-342=0\)

\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)

\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)

__________________

\(x\left(x+1\right)=650\)

\(\Rightarrow x^2+x=650\)

\(\Rightarrow x^2-x+650=0\)

\(\Rightarrow x^2+26x-25x-650=0\)

\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)

\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)

______________________

\(x\left(x+1\right)=380\)

\(\Rightarrow x^2+x=380\)

\(\Rightarrow x^2+x-380=0\)

\(\Rightarrow x^2+20x-19x-380=0\)

\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)

\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)

a, \(x\).(\(x\) + 1) = 156

156 = 2².3.13 = 12.13

Vậy \(x\).(\(x\) + 1) = 12.13

Vậy \(x\) = 12

b, \(x.\)(\(x\) + 1) =  342 

    342 = 2.3².19 = 18.19

    \(x\).(\(x+1\)) = 18.19

    \(x\) =   18

c, \(x\).(\(x\) + 1) = 650 

    650 = 2.5².13 = 25.26

    \(x\).(\(x\) +1) = 25.26

    \(x\) = 25

d, \(x\).(\(x\) +1) = 380 

    380 = 2².5.19 = 19.20

      \(x\).(\(x\) + 1) = 19.20

      \(x\)    = 19

chả lời đầy đủ nha

câu 1 x=97
câu 2 x=53