Question

Tìm x :

\(\dfrac{1}{1.2}.\left|2x-1\right|+\dfrac{1}{2.3}.\left|2x-1\right|+\dfrac{1}{3.4}+..+\dfrac{1}{1996.1997}.\left|2x-1\right|=1996\)

Answer 1

|2x - 1|.\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1996.1997}\right)\)= 1996

|2x - 1|.\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1996}-\dfrac{1}{1997}\right)\)= 1996

|2x - 1|.\(\left(1-\dfrac{1}{1997}\right)\)= 1996

|2x - 1|. ​\(\dfrac{1996}{1997}\)= 1996​​

$|2x\; -\; 1$| = 1996 : \(\dfrac{1996}{1997}\)

|2x - 1| = 1996 . \(\dfrac{1997}{1996}\)

|2x - 1| = 1997

2x - 1 = ± 1997

TH1:

2x -1 = 1997

2x = 1997 +1

2x= 1998

x= 1998:2

x=999

TH2:

2x-1= -1997

2x= -1997+1

2x= -1996

x= -1996:2

x= -998

Vậy x ∈ {999; -998}

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Answer 2

Phân phối phép nhân với phép cộng: v