Bài 1: Tìm x , biết :

a) ( x -1).(x-2)=0

<x-1=0

|

<x=0+1=1

-<x-2=0

-<x=0+2=2

Vậy x E {1;2}

b) (x-2).(x^2+1)=0

[<x-2=0

[<x=0+2=2

[>x²+1=0

   x²=0-1

   x²=1.(-1)

c) (x+`1).(x^2-4)=0

a) (x-5).(x+6)=0

=> x-5=0 hoặc x+6=0

Nếu x-5=0 thì x=0+5=5

Nếu x+6=0=>x=0-6=-6

vậy x=5 hoặc x=-6

b) |x|<4=>|x|=0;1;2;3=>x=0;1;-1;2;-2;3;-3

c) (x-7).(x+1)<0

=> x-7 và x+1 là hai số nguyên trái dấu

Vì x-7<x+1 nên x-7<0, x+1>0

Ta có:

x-7<0=>x<7

x+1>0=>x>-1

=> -1<x<7=> x=0;1;2;3;4;5;6

\(\frac{13}{14}+\frac{13}{35}+\frac{13}{65}+...+\frac{13}{1274}-x+2\frac{1}{5}=0\)

\(\frac{26}{28}+\frac{26}{70}+\frac{26}{130}+...+\frac{26}{2548}-x+2\frac{1}{5}=0\)

\(\frac{26}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}\right)-x+2\frac{1}{5}=0\)

\(\frac{26}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{49}-\frac{1}{52}\right)-x+2\frac{1}{5}=0\)

\(\frac{26}{3}.\left(\frac{1}{4}-\frac{1}{52}\right)-x+2\frac{1}{5}=0\)

\(\Rightarrow\frac{21}{5}-x=0\rightarrow x=\frac{21}{5}\)

\(\Leftrightarrow\left(x-5\right)^{x+1}\left[1-\left(x-5\right)^{12}\right]=0\)

=>(x-5)(1-x+5)(1+x-5)=0

=>(x-5)(6-x)(-4+x)=0

hay \(x\in\left\{5;6;4\right\}\)

a) (x-5).(x+6)=0 khi:

TH1: x-5=0 => x=5

TH2: x+6=0 => x=-6

Vậy x=5; x= -6.

b) |x|<4 =>x \(\in\)(1;2;3)

c)(x-7).(x+1)<0 khi:

TH1: x-7>0 và x+1<0 => x>7 và x<-1 => x\(\in\)\(\phi\)

TH2: x-7<0 và x+1>0 => x<7 và x<-1 => x<-1

Vậy x<-1.

d)  |2x-5|=13

TH1: 2x-5 =13=> 2x=18 => x=9

TH2: 2x-5 =-13 => 2x=-8 => x=-4

Vậy x=9; x=-4.

e)  |7x+3|=66

TH1: 7x+3=66 =>7x=63 => x=9

TH2:7x+3=-66 => 7x=-69 => x=-69:7. Mà -69 không chia hết cho 7=> x không có giá trị (vì đề ra x thuộc Z)

Vậy x=9.

f) |5x -2|≤13

TH1: 5x-2<13 => 5x<15 => x<3

TH2: 5x-2=13 => 5x=15 => x=3

Vậy x\(\le\)3.

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}