Question

Câu 2. (3 điểm). Tìm x biết:

a) $\dfrac{3}{8}-\dfrac{1}{6}x=\dfrac{1}{4}$;                 

b) ${{\left( x-1 \right)}^{2}}=\dfrac{1}{4}$;                  

c) $\left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0$.

Answer 1

A. 3/5 

B. 2/8 

C. 2 

HT

Answer 2

a) \(\frac{3}{8}-\frac{1}{6}x=\frac{1}{4}\)

\(\frac{1}{6}x=\frac{3}{8}-\frac{1}{4}\)

\(\frac{1}{6}x=\frac{1}{8}\)

\(x=\frac{1}{8}\div\frac{1}{6}\)

\(x=\frac{3}{4}\)

Vậy \(x=\frac{3}{4}\)

b) \(\left(x-1\right)^2=\frac{1}{4}\)

\(\left(x-1\right)^2=\left(\pm\frac{2}{4}\right)^2\)

TH1: 

\(\left(x-1\right)^2=\frac{2}{4}^2\)

\(x-1=\frac{2}{4}\)

\(x-1=\frac{1}{2}\)

\(x=\frac{1}{2}+1\)

\(x=\frac{3}{2}\)

TH2:

\(\left(x-1\right)^2=\left(-\frac{2}{4}\right)^2\)

\(x-1=-\frac{2}{4}\)

\(x=-\frac{2}{4}+1\)

\(x=\frac{1}{2}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

c) \(\left(x-\frac{-1}{2}\right).\left(x+\frac{1}{3}\right)=0\)

TH1:

\(x-\frac{-1}{2}=0\)

\(x=0+\frac{-1}{2}\)

\(x=\frac{-1}{2}\)

TH2:

\(x+\frac{1}{3}=0\)

\(x=0-\frac{1}{3}\)

\(x=-\frac{1}{3}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=-\frac{1}{3}\end{cases}}\)

Answer 3

=14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
$x$	=18:16=18:16
$x$	=34=34

x=34x=34.

(x−1)2=14(x−1)2=14                               

⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=32x=12[x=32x=12

x∈{32;12}x∈{32;12}.

(x−−12).(x+13)=0(x−−12).(x+13)=0.

⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

x∈{−12;−13}x∈{−12;−13}.

Answer 4

^3/8-1/6x=1/4

^1/6x=3/8-1/4

^1/6x=1/8

    ^x=1/8:1/6

    ^x=3/4

 

Answer 5

a, 3/8 - 1/6x = 1/4 

            1/6x = 3/8 - 2/8 

             1/6x = 1/8

                 x = 1/8 : 1/6

                 x = 3/4 

   vậy x = 3/4

b, ( x - 1 )^2 = 1/4 

suy ra : ( x -1 )^2 = ( 1/2 )^2

                    x - 1 = 1/2

                         x = 1/2 + 1

suy ra : x= 3/2

 vậy x = 3/2

c,( x - -1/2 ) . x 

Answer 6

Answer 7

Answer 8

a) \dfrac{3}{8}-\dfrac{1}{6}x83​−61​x =\dfrac{1}{4}=41​          

\dfrac{3}{8}-\dfrac{1}{6}x83​−61​x	=\dfrac{1}{4}=41​
\dfrac{1}{6}x61​x	=\dfrac{3}{8}-\dfrac{2}{8}=83​−82​
\dfrac{1}{6}x61​x	=\dfrac{1}{8}=81​
$x$	=\dfrac{1}{8}:\dfrac{1}{6}=81​:61​
$x$	=\dfrac{3}{4}=43​

Vậy x=\dfrac{3}{4}x=43​.

b) {{\left( x-1 \right)}^{2}}=\dfrac{1}{4}(x−1)2=41​                               

Suy ra \left[ \begin{aligned} &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\\ &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}\\ \end{aligned} \right.⎣⎢⎢⎢⎢⎡​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay \left[ \begin{aligned} &x-1=\dfrac{1}{2}  \\ &x-1=\dfrac{-1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x−1=21​ x−1=2−1​ ​ 

\left[ \begin{aligned} &x=\dfrac{1}{2}+1  \\ &x=\dfrac{-1}{2}+1  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=21​+1 x=2−1​+1 ​ suy ra \left[ \begin{aligned} &x=\dfrac{3}{2}  \\ &x=\dfrac{1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=23​ x=21​ ​

Vậy x\in \left\{ \dfrac{3}{2};\dfrac{1}{2} \right\}x∈{23​;21​}.

c) \left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0(x−2−1​).(x+31​)=0.

Suy ra \left[ \begin{aligned} &x-\dfrac{-1}{2} = 0\\ &x+\dfrac{1}{3} = 0\\ \end{aligned} \right.⎣⎢⎢⎡​​x−2−1​=0x+31​=0​  hay \left[ \begin{aligned} &x=\dfrac{-1}{2}  \\ &x=\dfrac{-1}{3}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=2−1​ x=3−1​ ​       

Vậy x\in \left\{ \dfrac{-1}{2};\dfrac{-1}{3} \right\}x∈{2−1​;3−1​

Answer 9

a) \dfrac{3}{8}-\dfrac{1}{6}x83​−61​x =\dfrac{1}{4}=41​          

\dfrac{3}{8}-\dfrac{1}{6}x83​−61​x	=\dfrac{1}{4}=41​
\dfrac{1}{6}x61​x	=\dfrac{3}{8}-\dfrac{2}{8}=83​−82​
\dfrac{1}{6}x61​x	=\dfrac{1}{8}=81​
$x$	=\dfrac{1}{8}:\dfrac{1}{6}=81​:61​
$x$	=\dfrac{3}{4}=43​

Vậy x=\dfrac{3}{4}x=43​.

b) {{\left( x-1 \right)}^{2}}=\dfrac{1}{4}(x−1)2=41​                               

Suy ra \left[ \begin{aligned} &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\\ &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}\\ \end{aligned} \right.⎣⎢⎢⎢⎢⎡​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay \left[ \begin{aligned} &x-1=\dfrac{1}{2}  \\ &x-1=\dfrac{-1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x−1=21​ x−1=2−1​ ​ 

\left[ \begin{aligned} &x=\dfrac{1}{2}+1  \\ &x=\dfrac{-1}{2}+1  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=21​+1 x=2−1​+1 ​ suy ra \left[ \begin{aligned} &x=\dfrac{3}{2}  \\ &x=\dfrac{1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=23​ x=21​ ​

Vậy x\in \left\{ \dfrac{3}{2};\dfrac{1}{2} \right\}x∈{23​;21​}.

c) \left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0(x−2−1​).(x+31​)=0.

Suy ra \left[ \begin{aligned} &x-\dfrac{-1}{2} = 0\\ &x+\dfrac{1}{3} = 0\\ \end{aligned} \right.⎣⎢⎢⎡​​x−2−1​=0x+31​=0​  hay \left[ \begin{aligned} &x=\dfrac{-1}{2}  \\ &x=\dfrac{-1}{3}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=2−1​ x=3−1​ ​       

Vậy x\in \left\{ \dfrac{-1}{2};\dfrac{-1}{3} \right\}x∈{2−1​;3−1​}.

Answer 10

a) \dfrac{3}{8}-\dfrac{1}{6}x83​−61​x =\dfrac{1}{4}=41​          

\dfrac{3}{8}-\dfrac{1}{6}x83​−61​x	=\dfrac{1}{4}=41​
\dfrac{1}{6}x61​x	=\dfrac{3}{8}-\dfrac{2}{8}=83​−82​
\dfrac{1}{6}x61​x	=\dfrac{1}{8}=81​
$x$	=\dfrac{1}{8}:\dfrac{1}{6}=81​:61​
$x$	=\dfrac{3}{4}=43​

Vậy x=\dfrac{3}{4}x=43​.

b) {{\left( x-1 \right)}^{2}}=\dfrac{1}{4}(x−1)2=41​                               

Suy ra \left[ \begin{aligned} &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\\ &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}\\ \end{aligned} \right.⎣⎢⎢⎢⎢⎡​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay \left[ \begin{aligned} &x-1=\dfrac{1}{2}  \\ &x-1=\dfrac{-1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x−1=21​ x−1=2−1​ ​ 

\left[ \begin{aligned} &x=\dfrac{1}{2}+1  \\ &x=\dfrac{-1}{2}+1  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=21​+1 x=2−1​+1 ​ suy ra \left[ \begin{aligned} &x=\dfrac{3}{2}  \\ &x=\dfrac{1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=23​ x=21​ ​

Vậy x\in \left\{ \dfrac{3}{2};\dfrac{1}{2} \right\}x∈{23​;21​}.

c) \left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0(x−2−1​).(x+31​)=0.

Suy ra \left[ \begin{aligned} &x-\dfrac{-1}{2} = 0\\ &x+\dfrac{1}{3} = 0\\ \end{aligned} \right.⎣⎢⎢⎡​​x−2−1​=0x+31​=0​  hay \left[ \begin{aligned} &x=\dfrac{-1}{2}  \\ &x=\dfrac{-1}{3}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=2−1​ x=3−1​ ​       

Vậy x\in \left\{ \dfrac{-1}{2};\dfrac{-1}{3} \right\}x∈{2−1​;3−1​}.

Answer 11

a) \dfrac{3}{8}-\dfrac{1}{6}x83​−61​x =\dfrac{1}{4}=41​          

\dfrac{3}{8}-\dfrac{1}{6}x83​−61​x	=\dfrac{1}{4}=41​
\dfrac{1}{6}x61​x	=\dfrac{3}{8}-\dfrac{2}{8}=83​−82​
\dfrac{1}{6}x61​x	=\dfrac{1}{8}=81​
$x$	=\dfrac{1}{8}:\dfrac{1}{6}=81​:61​
$x$	=\dfrac{3}{4}=43​

Vậy x=\dfrac{3}{4}x=43​.

b) {{\left( x-1 \right)}^{2}}=\dfrac{1}{4}(x−1)2=41​                               

Suy ra \left[ \begin{aligned} &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\\ &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}\\ \end{aligned} \right.⎣⎢⎢⎢⎢⎡​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay \left[ \begin{aligned} &x-1=\dfrac{1}{2}  \\ &x-1=\dfrac{-1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x−1=21​ x−1=2−1​ ​ 

\left[ \begin{aligned} &x=\dfrac{1}{2}+1  \\ &x=\dfrac{-1}{2}+1  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=21​+1 x=2−1​+1 ​ suy ra \left[ \begin{aligned} &x=\dfrac{3}{2}  \\ &x=\dfrac{1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=23​ x=21​ ​

Vậy x\in \left\{ \dfrac{3}{2};\dfrac{1}{2} \right\}x∈{23​;21​}.

c) \left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0(x−2−1​).(x+31​)=0.

Suy ra \left[ \begin{aligned} &x-\dfrac{-1}{2} = 0\\ &x+\dfrac{1}{3} = 0\\ \end{aligned} \right.⎣⎢⎢⎡​​x−2−1​=0x+31​=0​  hay \left[ \begin{aligned} &x=\dfrac{-1}{2}  \\ &x=\dfrac{-1}{3}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=2−1​ x=3−1​ ​       

Vậy x\in \left\{ \dfrac{-1}{2};\dfrac{-1}{3} \right\}x∈{2−1​;3−1​}.

Answer 12

A,3/4

B,3/2,1/2

C,-1/2,-1/3

Answer 13

a) \dfrac{3}{8}-\dfrac{1}{6}x83​−61​x =\dfrac{1}{4}=41​          

\dfrac{3}{8}-\dfrac{1}{6}x83​−61​x	=\dfrac{1}{4}=41​
\dfrac{1}{6}x61​x	=\dfrac{3}{8}-\dfrac{2}{8}=83​−82​
\dfrac{1}{6}x61​x	=\dfrac{1}{8}=81​
$x$	=\dfrac{1}{8}:\dfrac{1}{6}=81​:61​
$x$	=\dfrac{3}{4}=43​

Vậy x=\dfrac{3}{4}x=43​.

b) {{\left( x-1 \right)}^{2}}=\dfrac{1}{4}(x−1)2=41​                               

Suy ra \left[ \begin{aligned} &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\\ &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}\\ \end{aligned} \right.⎣⎢⎢⎢⎢⎡​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay \left[ \begin{aligned} &x-1=\dfrac{1}{2}  \\ &x-1=\dfrac{-1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x−1=21​ x−1=2−1​ ​ 

\left[ \begin{aligned} &x=\dfrac{1}{2}+1  \\ &x=\dfrac{-1}{2}+1  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=21​+1 x=2−1​+1 ​ suy ra \left[ \begin{aligned} &x=\dfrac{3}{2}  \\ &x=\dfrac{1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=23​ x=21​ ​

Vậy x\in \left\{ \dfrac{3}{2};\dfrac{1}{2} \right\}x∈{23​;21​}.

c) \left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0(x−2−1​).(x+31​)=0.

Suy ra \left[ \begin{aligned} &x-\dfrac{-1}{2} = 0\\ &x+\dfrac{1}{3} = 0\\ \end{aligned} \right.⎣⎢⎢⎡​​x−2−1​=0x+31​=0​  hay \left[ \begin{aligned} &x=\dfrac{-1}{2}  \\ &x=\dfrac{-1}{3}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=2−1​ x=3−1​ ​       

Vậy x\in \left\{ \dfrac{-1}{2};\dfrac{-1}{3} \right\}x∈{2−1​;3−1​}.

Answer 14

 

 

Answer 15

 

 

Answer 16

Answer 17

  

Answer 18

 

 

Answer 19

a,8/3-1/6x=1/4                                                                                                                              

        1/6x=8/3-1/4

           1/6x=32/12-3/12

            1/6x=29/12

                x=29/12:1//6

                x=29/12.6/1

                x=29/2

             vậy x=29/2

b, (x-1)²=1/4

      x-1²=1/4

      x-1=1/4

      x=1/4+1

       x= 1/4+4/4

       x=5/4

c,(x--1/2).(x+1/3)=0

   (x+1/2).(x+1/3)=0

   x+1/2.x+1/3=0

   x.(1/2+1/3)=0

   x.1/2+1/3=0

          x.1/2=0+1/3

          x.1/2=1/3

              x=1/3:1/2

              x=1/3.2/1

              x=3/2

           vậy x=3/2  

 

Answer 20

    

a) \dfrac{3}{8}-\dfrac{1}{6}x83​−61​x =\dfrac{1}{4}=41​          

\dfrac{3}{8}-\dfrac{1}{6}x83​−61​x	=\dfrac{1}{4}=41​
\dfrac{1}{6}x61​x	=\dfrac{3}{8}-\dfrac{2}{8}=83​−82​
\dfrac{1}{6}x61​x	=\dfrac{1}{8}=81​
$x$	=\dfrac{1}{8}:\dfrac{1}{6}=81​:61​
$x$	=\dfrac{3}{4}=43​

Vậy x=\dfrac{3}{4}x=43​.

b) {{\left( x-1 \right)}^{2}}=\dfrac{1}{4}(x−1)2=41​                               

Suy ra \left[ \begin{aligned} &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\\ &{{\left( x-1 \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}\\ \end{aligned} \right.⎣⎢⎢⎢⎢⎡​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay \left[ \begin{aligned} &x-1=\dfrac{1}{2}  \\ &x-1=\dfrac{-1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x−1=21​ x−1=2−1​ ​ 

\left[ \begin{aligned} &x=\dfrac{1}{2}+1  \\ &x=\dfrac{-1}{2}+1  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=21​+1 x=2−1​+1 ​ suy ra \left[ \begin{aligned} &x=\dfrac{3}{2}  \\ &x=\dfrac{1}{2}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=23​ x=21​ ​

Vậy x\in \left\{ \dfrac{3}{2};\dfrac{1}{2} \right\}x∈{23​;21​}.

c) \left( x-\dfrac{-1}{2} \right).\left( x+\dfrac{1}{3} \right)=0(x−2−1​).(x+31​)=0.

Suy ra \left[ \begin{aligned} &x-\dfrac{-1}{2} = 0\\ &x+\dfrac{1}{3} = 0\\ \end{aligned} \right.⎣⎢⎢⎡​​x−2−1​=0x+31​=0​  hay \left[ \begin{aligned} &x=\dfrac{-1}{2}  \\ &x=\dfrac{-1}{3}  \\ \end{aligned} \right.⎣⎢⎢⎡​​x=2−1​ x=3−1​ ​       

Vậy x\in \left\{ \dfrac{-1}{2};\dfrac{-1}{3} \right\}x∈{2−1​;3−1​}.

 

Answer 21

Answer 22

=14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
$x$	=18:16=18:16
$x$	=34=34

x=34x=34.

(x−1)2=14(x−1)2=14                               

⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=32x=12[x=32x=12

x∈{32;12}x∈{32;12}.

(x−−12).(x+13)=0(x−−12).(x+13)=0.

⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

x∈{−12;−13}x∈{−12;−13}.

Answer 23

=14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
$x$	=18:16=18:16
$x$	=34=34

x=34x=34.

(x−1)2=14(x−1)2=14                               

⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=32x=12[x=32x=12

x∈{32;12}x∈{32;12}.

(x−−12).(x+13)=0(x−−12).(x+13)=0.

⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

x∈{−12;−13}x∈{−12;−13}.

Answer 24

=14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
$x$	=18:16=18:16
$x$	=34=34

x=34x=34.

(x−1)2=14(x−1)2=14                               

⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=32x=12[x=32x=12

x∈{32;12}x∈{32;12}.

(x−−12).(x+13)=0(x−−12).(x+13)=0.

⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

x∈{−12;−13}x∈{−12;−13}.

Answer 25

=14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
$x$	=18:16=18:16
$x$	=34=34

x=34x=34.

(x−1)2=14(x−1)2=14                               

⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=32x=12[x=32x=12

x∈{32;12}x∈{32;12}.

(x−−12).(x+13)=0(x−−12).(x+13)=0.

⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

x∈{−12;−13}x∈{−12;−13}.

Answer 26

a) 3/5

b) 2/8 

c) 2 

Answer 27

=14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
$x$	=18:16=18:16
$x$	=34=34

x=34x=34.

(x−1)2=14(x−1)2=14                               

⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=32x=12[x=32x=12

x∈{32;12}x∈{32;12}.

(x−−12).(x+13)=0(x−−12).(x+13)=0.

⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

x∈{−12;−13}x∈{−12;−13}.

Answer 28

a) 38−16x38−16x =14=14          

38−16x38−16x	=14=14
16x16x	=38−28=38−28
16x16x	=18=18
xx	=18:16=18:16
xx	=34=34

Vậy x=34x=34.

b) (x−1)2=14(x−1)2=14                               

Suy ra ⎡⎢ ⎢ ⎢ ⎢⎣(x−1)2=(12)2(x−1)2=(−12)2[(x−1)2=(12)2(x−1)2=(−12)2 hay ⎡⎢ ⎢⎣x−1=12x−1=−12[x−1=12x−1=−12 

⎡⎢ ⎢⎣x=12+1x=−12+1[x=12+1x=−12+1 suy ra ⎡⎢ ⎢⎣x=32x=12[x=32x=12

Vậy x∈{32;12}x∈{32;12}.

c) (x−−12).(x+13)=0(x−−12).(x+13)=0.

Suy ra ⎡⎢ ⎢⎣x−−12=0x+13=0[x−−12=0x+13=0  hay ⎡⎢ ⎢⎣x=−12x=−13[x=−12x=−13       

Vậy x∈{−12;−13}x∈{−12;−13}

 

Answer 29

a) 3/8-1/6x= 1/4                       b)\(^{ }\left(x-1\right)^2\)=1/4                    c)(x—1/2) .(x+1/3)= 0

1/6x = 3/8-1/4                           x—1 = 1/4                          x.(1/2+1/3)=0

1/6x=1/8                                   x=1/4+1                              x.5/6 =0

x=1/6:1/8                                  x=5/4                                  x=0.5/6

x=6/8                                                                                  x=0

Answer 30

a,6/48

b,4/4

c,0