a) x+7=-12

        x=(-12)-7

        x=-19

b)x-15=-21

        x=(-21)+15

        x=-6

c)13-x=20

        x=13-20

        x=-7

d)17-(2+x)=3

      x=17-3

      x=14

      x=14-2

      x=12

a,x+7=-12

=>x= -12-7

=>x= -19

b,x-15= -21

=>x= -21+15

=>x= -6

c,13-x=20

=>x=13-20

=>x= -7

d, 17-(2+x)=3

=>2+x=17-3

=>2+x=14

=>x=14-2

=>x=12

a) \(x+7=-12\) 

    \(x=-12-7\) 

    \(x=-19\) 

b) \(x-15=-21\)  

    \(x=-21+15\) 

    \(x=-6\)  

c) \(13-x=20\) 

            \(x=13-20\)  

            \(x=-7\) 

d) \(17-\left(2+x\right)=3\)  

             \(2+x=17-3\)  

             \(2+x=14\)   

                   \(x=14-2\)  

                   \(x=12\)

a/ x - 7 = 12

=> x = 12 + 7 = 19

b/ 9 + 4(x - 5) = 13

=> 4x - 20 = 4

=> 4x = 24

=> x = 6

c/ (x+2)³ = 64

=> x + 2 = 4

=> x = 2

a) x-7=12

     x=12+7

     x=19

b) 9+4.(x-5)=13

       4.(x-5)=13-9

       4.(x-5)=4

          (x-5)=4:4

          (x-5)=1

          x=5+1

            x=6

cau cuoi mik ko bt lam nha!

chuc ban hoc tot

a/ x - 7 = 12

=> x = 12 + 7 = 19

b/ 9 + 4(x - 5) = 13

=> 4x - 20 = 4

=> 4x = 24

=> x = 6

c/ (x+2)³ = 64

=> x + 2 = 4

=> x = 2

a; Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x-y}{7-4}=\dfrac{12}{3}=4\)

Do đó: x=28; y=16

\(a,\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x-y}{7-4}=\dfrac{12}{3}=4\\ \Rightarrow\left\{{}\begin{matrix}x=4.7=28\\y=4.4=16\end{matrix}\right.\\ b,\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{2}=\dfrac{z}{2}\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{6}=\dfrac{x+y+z}{4+6+6}=\dfrac{50}{16}=\dfrac{25}{8}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{25}{8}.4=\dfrac{25}{2}\\y=\dfrac{25}{8}.6=\dfrac{75}{4}\\z=\dfrac{25}{8}.6=\dfrac{75}{4}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+14x+49-x^2+3x=12\\ \Leftrightarrow17x=-37\Leftrightarrow x=-\dfrac{37}{17}\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^2+2x7+49-x^2+3x=12\Leftrightarrow17x=-37\Leftrightarrow x=\dfrac{-37}{17}\)

b) \(x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=\left(0\right)\Leftrightarrow x=1,x=2\)

`a,`

`(x+7)^2-x(x-3)=12`

`<=>x^2+14x+49 - x^2 +3x=12`

`<=> 17x =-37`

`<=>x=(-37)/17`

Vậy `x=(-37)/17`

`b,`

`x^2-3x+2=0`

`<=>x^2-2x-x+2=0`

`<=>x(x-2)-(x-2)=0`

`<=>(x-2)(x-1)=0`

giải ra `x=2,x=1`

Vậy `x=2,x=1`

a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\)

<=> \(x=\dfrac{7}{12}-\dfrac{5}{24}=\dfrac{14}{24}-\dfrac{5}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\)

<=> \(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)

c) bn ghi rõ đề chút

ok bạn

(2\7.x+3\7) :2 1\5-3\7=1

hai phần bảy nhân x cộng ba phần bảy chia hai một phần năm trừ 3 phần bảy bằng một

Giải:

a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\) 

              \(x=\dfrac{7}{12}-\dfrac{5}{24}\) 

              \(x=\dfrac{3}{8}\) 

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\) 

            \(x=\dfrac{1}{2}+\dfrac{3}{4}\) 

            \(x=\dfrac{5}{4}\) 

c) \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):2\dfrac{1}{5}-\dfrac{3}{7}=1\) 

           \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):\dfrac{11}{5}=1+\dfrac{3}{7}\) 

           \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):\dfrac{11}{5}=\dfrac{10}{7}\) 

                        \(\dfrac{2}{7}.x+\dfrac{3}{7}=\dfrac{10}{7}.\dfrac{11}{5}\) 

                        \(\dfrac{2}{7}.x+\dfrac{3}{7}=\dfrac{22}{7}\)           

                                \(\dfrac{2}{7}.x=\dfrac{22}{7}-\dfrac{3}{7}\) 

                                \(\dfrac{2}{7}.x=\dfrac{19}{7}\) 

                                    \(x=\dfrac{19}{7}:\dfrac{2}{7}\) 

                                    \(x=\dfrac{19}{2}\)

\(2x^3+x^2-4x-12\)

\(=2x^3+5x^2+6x-4x^2-10x-12\)

\(=\left(2x^3+5x^2+6x\right)-\left(4x^2+10x+12\right)\)

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

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

\(a,2x^3+x^2-4x-12=\left(2x^3-4x^2\right)+\left(5x^2-10x\right)+\left(6x-12\right)=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)

\(b,x^5-xy^4+x^4y-y^5=x\left(x^4-y^4\right)+y\left(x^4-y^4\right)=\left(x+y\right)\left(x^4-y^4\right)=\left(x+y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)^2\left(x-y\right)\left(x^2+y^2\right)\)

\(c,\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)-9=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]-9=\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9\)

đặt \(x^2+8x+11=y\)

\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9=\left(y-4\right)\left(y+4\right)-9=y^2-16-9=y^2-25=\left(y-5\right)\left(y+5\right)=\left(x^2+8x+11-5\right)\left(x^2+8x+11+5\right)=\left(x^2+8x+6\right)\left(x^2+8x+16\right)=\left(x^2+8x+6\right)\left(x+4\right)^2\)

Lời giải:
a. Đề có cả x,y. Bạn xem lại

b. 

PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$

$\Leftrightarrow (x-3)(5x-2)=0$

$\Leftrightarrow x-3=0$ hoặc $5x-2=0$

$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$

c.

PT $\Leftrightarrow (7x-2)(x-4)=0$

$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$

$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$

d. Đề thiếu.