Chia nhỏ ra ik ạ

\(\left(2x+3\right)\left(2x-3\right)-4x\left(x+5\right)=4x^2-9-4x^2-20x=-20x-9\)

\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)

\(x\left(x+2\right)-\left(x-3\right)\left(x+3\right)=x^2+2x-\left(x^2-9\right)=x^2+2x-x^2+9=2x+9\)

1: Ta có: \(20-2\left(x+4\right)=4\)

\(\Leftrightarrow2\left(x+4\right)=16\)

\(\Leftrightarrow x+4=8\)

hay x=4

5: Ta có: \(\left(x+1\right)^3=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

Bài này giống tìm nghiệm quá :

1) \(5\left(x-7\right)=0\)

    \(\left(x-7\right)=0\div5\)

       \(\left(x-7\right)=0\)

     \(x=0+7\)

       \(x=7\)

2) \(25\left(x-4\right)=0\)

          \(\left(x-4\right)=0\div25\)

           \(\left(x-4\right)=0\)

              \(x=0+4\)

              \(x=4\)

3) \(34\left(2x-6\right)=0\)

    \(\left(2x-6\right)=0\div34\)

      \(\left(2x-6\right)=0\)

        \(2x=0+6\)

          \(2x=6\)

           \(x=6\div2\)

          \(x=3\)

4) \(2007\left(3x-12\right)=0\)

               \(\left(3x-12\right)=0\div2007\)

                 \(\left(3x-12\right)=0\)

                   \(3x=0+12\)

                    \(3x=12\)

                     \(x=12\div3\)

                       \(x=4\)

5) \(47\left(5x-15\right)=0\)

         \(\left(5x-15\right)=0\div47\)

           \(\left(5x-15\right)=0\)

            \(5x=0+15\)

              \(5x=15\)

                 \(x=15\div5\)

                 \(x=3\)

6) \(13\left(4x-24\right)=0\)

         \(\left(4x-24\right)=0\div13\)

          \(\left(4x-24\right)=0\)

             \(4x=0+24\)

            \(4x=24\)

              \(x=24\div4\)

                \(x=6\)

1, 2x - 35 = 15

  2x         =  15 + 35

  2x         =   50

    x         =  50 : 2 

    x         =   25.

2, 3x + 18 = 12

    3x        =  12 - 18

    3x        = -6

     x         =  -6 : 3

     x         =  -2.

3, / x - 1 / = 0

=> x \(\in\varnothing\).

4, -13 /x/ = - 26

  /x/        = -26 : -13

=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

 Vậy x \(\in\){ 2 ; -2}.

5,4 - ( 27 - 3 ) = x - ( 13 - 4 )

  4 - 24           = x - 9

  -20               =  x - 9

             -x      =    9 + 20

             -x      =    29

              x      =   -29.

6, 47 - ( x + 15 ) = 21

  47 - x - 15        = 21

       -x - 15         = 21 - 47 

       -x - 15         = -26

             -x          =  -26 + 15 

             -x          =     - 11

              x          = 11.

7, -5 -( 24 - x) = - 11

   -5 - 24 + x   = -11

        -24 + x   = -11 + 5

        -24 + x   =  -6

                 x   = -6 + 24

                 x   =  18.

8, 6 - /x/ = 2

        /x/   = 6 - 2

         \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy x \(\in\left\{3;-3\right\}.\)

9, 6 + /x/ = 2

         /x/ = 2 - 6

=>      x = -4.

2x - 35 = 15

=> 2x = 15 + 35

=> x = 50 : 2

=> x = 25

3x + 18 = 12

=> 3x = 12 - 18

=> x = ( -6 ) : 3

=> x = -2

| x - 1 | = 0

=> x - 1 = 0

=> x = 0 + 1

=> x = 1

-13 * | x | = -26

=> | x | = -26 : ( -13 )

=> | x | = 2

47 - (x + 5) = -6 - (24 - x)

<=> 47 - x - 5 = -6 - 24 + x

<=> - x - x = -6 - 24 - 47 + 5

<=> - 2x = -72

<=> x = 36

cảm ơn bạn rất nhiều

tìm y nhé

47 + (2 x + 5) = -6 - (24 - y)

47 + 10 = -6 - (24 - y)

57 = -6 - (24 - y)

57 + 6 = 24 - y

63 = 24 - y

y = 24 - 63

y = -39 nhé anh, em học lớp 5

47 + (2 x y + 5) = -6 - (24 - y)

chứ ko phải 

47 + (2 x  + 5) = -6 - (24 - y)

đó em

Bài 1: 

1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)

\(=-12-18\)

=-30

2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)

\(=36-2020+2019-136-27\)

\(=1-100-27\)

\(=-126\)

3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)

\(=144-97-244+197\)

\(=-100+100=0\)

4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)

\(=-24\cdot13+24\cdot3\)

\(=24\cdot\left(-13+3\right)\)

\(=24\cdot\left(-10\right)=-240\)

5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)

\(=54+55+56+57+58-64-65-66-67-68\)

\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)

\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)

=-50

6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)

\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)

\(=-24\cdot5+16\cdot5\)

\(=5\cdot\left(-24+16\right)\)

\(=-5\cdot8=-40\)

7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)

\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)

\(=47\cdot50-23\cdot50\)

\(=50\cdot\left(47-23\right)\)

\(=50\cdot24=1200\)

8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)

\(=-31\cdot\left(47+52+1\right)\)

\(=-31\cdot100=-3100\)

Bài 2: 

1) Ta có: \(-17-\left(2x-5\right)=-6\)

\(\Leftrightarrow-17-2x+5+6=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow-2x=6\)

hay x=-3

Vậy: x=-3

2) Ta có: \(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow10-8+6x+4=0\)

\(\Leftrightarrow6x+6=0\)

\(\Leftrightarrow6x=-6\)

hay x=-1

Vậy: x=-1

3) Ta có: \(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow-12-3x+21+18=0\)

\(\Leftrightarrow-3x+27=0\)

\(\Leftrightarrow-3x=-27\)

hay x=9

Vậy: x=9

4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)

\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)

\(\Leftrightarrow-2x-3=-3\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: x=0

5) Ta có: x(x+3)=0

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

Vậy: \(x\in\left\{0;-3\right\}\)

6) Ta có: (x-2)(x+4)=0

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

Vậy: \(x\in\left\{2;-4\right\}\)

7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)

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

Vậy: \(x\in\left\{0;-1;3\right\}\)

Bài 1: 

1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)

=−12−18=−12−18

=-30

2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27

=36−2020+2019−136−27=36−2020+2019−136−27

=1−100−27=1−100−27

=−126

Tớ chcs cậu học thật giỏi nha !