a, 10 + 2x = 45 ÷ 43

10 + 2x = 45/43

2x = 45/43 - 10

2x = -385/43

x = -385/43 ÷ 2

x = -385/86

b, 2x - 138 = 23.32

2x - 138 = 736

2x = 736 + 138

2x = 874

x = 874 ÷ 2

x = 437

c, 231 - ( x - 6 ) = 1339 ÷ 13

231 - ( x - 6 ) = 103

x - 6 = 231 - 103

x - 6 = 128

x = 128 + 6

x = 134

d, x + 5.2 - ( 32 + 16.3 ÷ 6 - 15 ) = 0

x + 10 - 25 = 0

x + 10 = 0 + 25

x + 10 = 25

x = 25 - 10

x = 15

a) 10 + 2 . x = 45 : 43

            2 . x = 45/43 -10

            2 . x = -385/43

                 x = -385/43 : 2

                 x = -385/86

b) 2 . x - 138 = 23 . 32

    2 . x - 138 = 736

             2 . x = 736+138

             2 . x = 874

                  x = 874 : 2

                  x = 437

c) 231 - (x -6) = 1339 : 13

    231 - (x -6) = 103

               x - 6 = 231 -103

               x - 6 = 98

                    x = 98 +6

                    x = 104

d) x + 5 . 2 - (32 +16 . 3 : 6 - 15) = 0

        x + 5 . 2 - (32 + 48 : 6 - 15) = 0

               x + 5 . 2 - (32 + 8 - 15) = 0

                               x + 5 . 2 - 25 = 0

                               x + 10 - 25 = 0

                                      x + 10 = 0 + 25

                                      x + 10 = 25

                                              x = 25 - 10

                                              x = 15

ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\)

\(=\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{\left(\sqrt{x}+3\right)^2}\)

\(=\dfrac{3\sqrt{x}+9-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)

\(=\dfrac{\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)

\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

Vậy...

a, =>-2x+12+6x-60=8

=>4x+48=8

=>4x=56

=>x=14

b,=>-8x-36-8x-3-x-13=0

=>-17x-52=0

=>-17x=52

=>x=-52/17

c,=>14x+7x^2-7x^2-21x=14

=>-7x=14

=>x=-2

a <=> -2x+12 +6x-60-8=0

<=>4x-56=0

<=>x=56/4=14

Vậy S=(4)

giải phá các ngoặc ra rút gọn x rồi tìm

ko ghi lại đề nha ! 

a) \(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+x\left(2^2-x^2\right)=0\)

\(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+4x-x^3=0\)

\(\Leftrightarrow-27+4x=0\)

\(\Leftrightarrow4x=27\)

\(\Leftrightarrow x=6,75\) 

b)\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=-0,5\)

a

\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

b

x^3 chứ: )

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

​a) | 2x - 5 | = 13

​=> 2x - 5 = 13 hoặc 2x - 5 = -13

​+ Nếu 2x - 5 = 13

2x = 13 + 5

2x = 18

x = 18 : 2

x = 9

​+ Nếu 2x - 5 = -13

​2x = ( -13 ) + 5

​2x = -8

​x = ( -8 ) : 2

x = -4

=> x = { -4 ; 9 }

​Tck nha

|7x + 3| = 66 

7x + 3 = 66 

7x       = 66-3

7x       = 63

 x        = 63 : 7

x         = 9

a. \(1-2x< 7\)

mà: \(1-n\le1\)với mọi n

\(\Rightarrow2x=n\Rightarrow x=\frac{n}{2}\)với mọi n

b.để: (x-1).(x-2)>0

=> x-1>0hoặc x-2<0

=>x>1hoặc x<2

(mik chỉ làm 2 câu mẫu thôi, bạn cố gắng tự làm nha, rất vui được kết bạn với bạn)

`1/9(x-3)^2-1/25(x+5)^2=0`

`<=>(1/3x-1)^2-(1/5x+1)^2=0`

`<=>(1/3x-1-1/5x-1)(1/3x-1+1/5x+1)=0`

`<=>(2/15x-2). 8/15x=0`

`<=>2/15x-2=0` hoặc `8/15x=0`

`<=>x=15`         hoặc `x=0`

Vậy `S=`{`15;0`}