a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

a,2.(3+X)=30-14

2.(3+X)=16

3+x=16/2

3+x=8

x=5

a)30-2(3+x)=14

        2(3+x)=30-14

        2(3+x)=16

           3+x =16:2

           3+x =8

               x =8-3

               x =5

b)15-3(10-x)=9

         3(10-x)=15-9

         3(10-x)=6

             10-x=6:3

             10-x=2

                  x=10-2

                  x=8

c)16-4(2x-9)=8

        4(2x-9)=16-8

        4(2x-9)=8

            2x-9=8:4

            2x-9=2

                2x=2+9

                2x=11

                 x= 11:2

                 x=5,5

 cho mk 1 nha

a) 30-2(3+x)=14

         2(3+x)=30-14

         2(3+x)=16

           (3+x)=16/2=8

              x=8-3=5

b) 15-3(10-x)=9

          3(10-x)=15-9

         3(10-x)=6

            10-x=6/3=2

                x=10-2=8

c) 16-4(2x-9)=8

         4(2x-9)=16-8=8

            2x-9=8/4=2

            2x   =2+9=11

             x=11/2=5,5

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

a.

\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)

\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)

\(18x-2=16\)

\(18x=16+2\)

\(18x=18\)

\(x=\frac{18}{18}\)

\(x=1\)

b.

\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(10x^2+9x-10x^2-15x+2x+3=8\)

\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)

\(-4x=8-3\)

\(-4x=5\)

\(x=-\frac{5}{4}\)

c.

\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)

\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)

\(42x=41\)

\(x=\frac{41}{42}\)

- mọi người ơi giúp mình với ạ. ai mình cx cho đúng

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

a)4(x+2)-7(2x-1)+9(3x-4)=30                                                     b)2(5x-8)-3(4x-5)=4(3x-4)+11

<=>4x+8-14x+7+27x-36=30                                            <=>10x-16-12x+15=12x-16+11

<=>17x-21=30                                                                    <=>  -14x=-4     <=>x=2/7

<=>17x=51

<=>x=3 

phần cvaf d đâu bạn ơi

phần c  hình như có vấn đề

d)(5x-3).4x-2x(10x-3)=15

<=>20x²-12x-20x²+6x=15

<=>   -6x=15

<=>x=-15/6

\(a,\Rightarrow x=3\)

\(b,\Rightarrow2x-1=2\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\dfrac{3}{2}\)

\(c,\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

\(d,\Rightarrow2x-3=3\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

a) x³ = 27 

x³ = 3³

x = 3

a: ĐKXĐ: x>=2/3

\(\dfrac{x-2}{\sqrt{3x-2}+2}=9\)

=>\(x-2=9\sqrt{3x-2}+18\)

=>\(9\sqrt{3x-2}=x-2-18=x-20\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=20\\81\left(3x-2\right)=x^2-40x+400\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=20\\x^2-40x+400-243x+162=0\end{matrix}\right.\)

=>x>=20 và x^2-283x+562=0

=>x=281(nhận) hoặc x=2(loại)

b: ĐKXĐ: x>=2/5

\(\sqrt{5x-2}=9\)

=>5x-2=81

=>5x=83

=>x=83/5

c: ĐKXĐ: x>=-1; x<>8

\(\dfrac{2x-16}{\sqrt{x+1}-3}=5\)

=>\(2x-16=5\sqrt{x+1}-15\)

=>\(\sqrt{25x+25}=2x-16+15=2x-1\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-4x+1=25x+25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-29x-24=0\end{matrix}\right.\)

=>x=8(nhận) hoặc x=-3/4(loại)