x.( 7x - 42 ) + 10. ( 7x - 42 ) = 0
=> ( 7x - 42 ). ( x + 10 ) = 0
=> \(\orbr{\begin{cases}x+10=0\\7x-42=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-10\\x=6\end{cases}}\)
Tìm x biết :
7x + 14 = 42
a) 23+x=150
b) 300 : x - 18 =42.
c) 7x+3 = 723.78.
a) 7x - 8 = 713
b) 8 (x - 3) = 0
c) 4x : 17 = 0
d) 96 - 3 ( x + 1 ) = 42
(x+5)-2(x-7)+5.(8+x)-7.3(3-2x)=(42-x)-5.(7x-7)+(4x-1)+3462
(x+5)-2.(x-7)+5.(8+x)-7.(3-2x)=(42-x)-5.(7x-7)+(4x-1)+34602
Tìm x nguyên
a) |6x-3| = 15
b) (x+7)(8-x) = 0
c) |7x-2| \(\le\)19
d) (2x-1)(4y-2)=-42
Tim tat ca cac so x sao cho:
(x^2-7x+11)^(x^2-13x+42)=1
a)142.42 - (7.52 + 4.x) = (43.157):42
b)[(7x + 23) : 62 - 522] :29 = 182
Tìm x:biết / / là giá trị tuyệt đối
3./x-1/=96
2x+42=-20
7x-12=37
