Tìm x, biết [(14x+26).3+55]:5=35

b) 1500 (x-7) =0

=> x -7 =0

=> x = 7

c) ( 2x-4)(48-12x) =0
TH1 :

2x-4 = 0

=> 2x = 4

=> x =2
TH2 :

45-12x =0

=> 12x =48

=> x =48/12

=> x = 4 

b, 1500.(x – 7) = 0

<=>1500x-10500=0

<=>1500x=10500

<=>x=7

Vậy x=7

c,(2.x – 4).(48 – 12.x) = 0

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-4=0\\48-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\12x=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy x=2 hoặc x=4

d, (x + 12).(x – 1) =0

\(\Leftrightarrow\left\{{}\begin{matrix}x+12=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-12\\x=1\end{matrix}\right.\)

Vậy x=-12 hoặc x=1

Bài 2:

a) 128- 3(x+ 4) = 23

\(\Leftrightarrow\)128-(3x+12)=23

\(\Leftrightarrow\)128-3x-12=23

\(\Leftrightarrow\)116-3x=23

\(\Leftrightarrow\)3x=116-23

\(\Leftrightarrow\)3x=93

\(\Leftrightarrow\)x=31

Vậy x=31

b) [(14x+ 26). 3+ 55]: 5= 35

\(\Leftrightarrow\)(14x+ 26). 3+ 55=175

\(\Leftrightarrow\)42x+78+55=175

\(\Leftrightarrow\)42x+133=175

\(\Leftrightarrow\)42x=175-133

\(\Leftrightarrow\)42x=42

\(\Leftrightarrow\)x=1

Vậy x=1

d, 720: [41- (2x- 5)]= 23. 5

\(\Leftrightarrow\)720: 41- (2x- 5)=115

\(\Leftrightarrow\)41-(2x- 5)=720:115

\(\Leftrightarrow\)41-(2x- 5)=\(\dfrac{144}{23}\)

\(\Leftrightarrow\)2x-5=\(\dfrac{799}{23}\)

\(\Leftrightarrow\)2x=\(\dfrac{914}{23}\)

\(\Leftrightarrow\)x=\(\dfrac{457}{23}\)

Vậy x=\(\dfrac{457}{23}\)