\(-4+x=-4+7\)

\(\Rightarrow x=-4+7+4\)

\(\Rightarrow x=7\)

\(\left(-3\right)-\left(-5-x\right)=-15+\left(-6+15\right)\)

\(\Rightarrow x=-15-6+15+3-5\)

\(\Rightarrow x=-8\)

\(-\left(x+2\right)+2=\left(8-15\right)+15\)

\(\Rightarrow-x+2+2=-7+15\)

\(\Rightarrow-x=-7+15-2-2\)

\(\Rightarrow x=-4\)

\(4+\left(x-4\right)=-\left(11-2\right)+\left(11+4\right)\)

\(\Rightarrow4+x-4=-11+2+11+4\)

\(\Rightarrow x=-11+2+11+4-4+4\)

\(\Rightarrow x=6\)

a) -4 + x = -4 + 7            b) (-3) - (-5-x) = -15 + (-6+15)

   -4 + x + 4 - 7 = 0            (-3) + 5 + x = -6

        x - 7 = 0                     2 + 6 = -x

        x= 7                            8 = -x 

 Vậy x= 7                            x= -8   =) Vậy x= -8

c) -(x+2)+2=(8-15)+15          d) 4+(x-4)= -(11-2) + (11+4 )

   -x - 2 + 2 = 8                       4+x-4 = 6

   -x = 8                                     x=6

    x= -8                              Vậy x=6

Vậy x= -8

Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

8/15 - 2/15 : x = 0,2

2/15 : x = 8/15 - 0,2

2/15 : x = 1/3

x = 2/15 : 1/3

x = 2/5

em làm ngược lại thoi 

2/15 : x = 8 /15 - 0,2

x = 2/15:( 8 /15 - 0,2)

x = 16

\(a,\left|15+x\right|+x=-15\)

\(\Rightarrow\left|15+x\right|=-15-x\)

\(\Rightarrow\left|15+x\right|=-\left(15+x\right)\)

Vì \(\left|15+x\right|\ge0\forall x;-\left(15+x\right)\le0\forall x\)

\(\Rightarrow15+x=-15-x=0\Rightarrow x=-15\)

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

a) x - 15 = 7 + (-2)

x-15 = 5

x=20

b) | x - 5 | = 15 + (-8)

I x - 5 I = 7

1) x-5 = 7 => x = 12

2) x-5 = -7 => x= -2

Vậy x=12 hoặc x=-2

c) I x | -  25 = 45 + (-15) 

 I x | -  25 = 30

I xI = 55

Vậy x= 55 hoặc x=-55

a) x - 15 = 7 + ( - 2 )

x - 15 = 5

      x  = 5 + 15

      x  = 20

b) | x - 5 | = 15 + ( - 8 )

| x - 5 | = 7

=> x = 7 + 5 

     x = 12

c) | x | - 25 = 45 + ( - 15 )

| x | - 25 = 30

=> | x | = 30 + 25

     | x | = 55

=> x = 55

phần cuối x = - 55 hoặc 55

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