minh moi hoc lop 5 thoi

1) 3x - 6=  5x + 2

5x - 3x = -6 - 2

2x = -8

x = -4

2) 15 - x = 4x - 5

4x + x = 15 + 5

5x = 20

x = 4

Tương tự như trên 

1) 3x - 6 = 5x + 2

  3x - 5x = 2 + 6

        -2x = 8

           x = 8 : (-2)

           x = -4

2) 15 - x = 4x - 5

    -x - 4x = -5 - 15

         -5x = -20

            x = -20 : (-5)

            x = 4

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

\(6.\left(x-2\right)+\left(-2\right)=20-4x\)

\(\Leftrightarrow\)\(6x-12-2=20-4x\)

\(\Leftrightarrow\)\(6x-4x=20+12+2\)

\(\Leftrightarrow\)\(2x=34\)

\(\Leftrightarrow\)\(x=\frac{34}{2}=17\)

Vậy \(x=17\)

6(x-2)+(-2)=20-4x

6x - 6.2 - 2 = 20 - 4x

6x - 12 - 2 = 20 - 4x

6x - 14 = 20 - 4x

6x + 4x = 20 + 14

10x = 34

x = 34 : 10

x = 3,4

6(x-2)+(-2)=20-4x

=> 6x-12-2=20-4x

=>.       6x+4x=20+12+2

=>.      10x=34

=>           x=34:10

=>.          x=34/10

         Vậy x= 34/10

Nhìu quá bạn ơi

nhiều vậy ai mà làm dc bạn ơi

làm hai câu ba câu cũng được

\(6\left(x-2\right)+\left(-2\right)=20-4x\)

\(\Leftrightarrow6x-12-2=20-4x\)

\(\Leftrightarrow6x+4x=20+12+2\)

\(\Leftrightarrow10x=34\)

\(\Leftrightarrow x=\frac{34}{10}=3,4\)

Vậy x= 3, 4

Học tốt !

từ pt đã cho ta có

6x-12-2=20-4x

<=>6x+4x=20+14

<=>10x=34

<=>x=3/4

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

a) x+15=20-4x

    x+4x=20-15

        5x=5

          x=5:5

          x=1

Vậy x=1

b) 8x+|-3|=-4x+39

    8x+4x=39-3

    12x=36

        x=36:12

        x=3

Vậy x=3

c) 6(x-2)+(-2)=20-4x

    6x-12-2=20-4x

    6x+4x=20+12+2

    10x=34

        x=34:10

        x=3,4

Vậy x=3,4

a) x+15=20-4x

<=> x+15-20+4x=0

<=> 5x+5=0

<=> 5x=-5

<=>x=-1

Vậy x=-1

b) 8x+|-3|=-4x+39

<=> 8x+3=-4x+39

<=> 8x+3+4x-39=0

<=> 12x-30=0

<=>12x=30

<=>x=5/2

Vậy x=5/2

c) 6(x-2)+(-2)=20-4x

<=> 6x-12-2=20-4x

<=> 6x-14=20-4x

<=> 6x-14-20+4x=0

<=> 10x-34=0

<=>10x=34

<=>x=17/5

Vậy x=17/5

a) x+15=20-4x

   x + 4x = 20 - 15

     5x      = 5

         x    = 5 : 5

         x     = 1

b) 8x+|-3|=-4x+39

   8x + 3   = -4  + 39

    8x + 3  = 35

    8x        = 35 - 3

       8x     = 32

       x       = 32 : 8

       x       = 4

c) 6(x-2)+(-2)=20-4x

    6x - 12 + (-2) = 20 - 4x

     6x + 4x         = 20 + 12 + 2

       10x              = 34

        x                 = 34 : 10

          x               = 3,4

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