a) \(x+215=480\)

\(x=480-215\)

\(x=265\)

b) \(725-x=185:5\)

\(725-x=37\)

\(x=725-37\)

\(x=688\)

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

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

\(x\times\dfrac{5}{3}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\times\dfrac{3}{5}\)

\(x=\dfrac{2}{5}\)

d) \(\left(\dfrac{13}{5}+\dfrac{2}{3}\right):x=5\)

\(\dfrac{49}{15}:x=5\)

\(x=\dfrac{49}{15}:5=\dfrac{49}{15}\times\dfrac{1}{5}\)

\(x=\dfrac{49}{75}\)

a)x + 215 = 480 

x = 480 - 215 

x = 265 

b) 725 - x = 185 : 5 

    725 - x = 37

              x = 725 - 37 

             x = 688

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

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

    \(x\) \(\times\) \(\dfrac{5}{3}\)          = \(\dfrac{2}{3}\)

    \(x\)                  = \(\dfrac{2}{3}\) : \(\dfrac{5}{3}\)

    \(x\)                  = \(\dfrac{2}{5}\)

d, (\(\dfrac{13}{5}\) + \(\dfrac{2}{3}\)) : \(x\) = 5

     \(\dfrac{49}{15}\) : \(x\)        = 5

             \(x\)           = \(\dfrac{49}{15}\) : 5

              \(x\)         = \(\dfrac{49}{75}\)

có thể viết rõ hơn ko mình ko hiểu

a) \(\Rightarrow x^2+4x+25-x^2=3\Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\)

b) \(\Rightarrow\left(4x+3-2x+3\right)\left(4x+3+2x-3\right)=0\)

\(\Rightarrow2\left(x+3\right).6x=0\)

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

Bài 1

2/5 - 3/5 : (0,75 + 30%)

= 2/5 - 3/5 : 21/20

= 2/5 - 4/7

= -6/35

‐--------------------

Bài 2

a) x/6 = -11/3

x = -11/3 . 6

x = -22

b) 16/5 của x là 48

x = 48 : 16/5

x = 15

bài1

2/5-3/5:(0,75+30%)=2/5-3/5:(75/100+30/100)

                               =2/5-3/5:105/100

                               =2/5-3/5:21/20

                               = 2/5-3/5. 20/21

                               =2/5-1/1.4/7

                               =2/5-4/7

                               =2/5+-4/7

                               =14/35+-20/35

                               =34/35

\(a,2x+\dfrac{5}{2}=-\dfrac{3}{5}\)

\(2x=-\dfrac{3}{5}-\dfrac{5}{2}\)

\(2x=-\dfrac{31}{10}\)

\(x=-\dfrac{31}{10}:2\)

\(x=-\dfrac{31}{20}\)

\(b,\dfrac{1}{2}:x-\dfrac{5}{6}=-\dfrac{2}{3}\)

\(\dfrac{1}{2}:x=-\dfrac{2}{3}+\dfrac{5}{6}\)

\(\dfrac{1}{2}:x=\dfrac{1}{6}\)

\(x=\dfrac{1}{2}:\dfrac{1}{6}\)

\(x=3\)

\(c,\left(312-x\right):12,6=24,5\)

\(312-x=24,5\times12,6\)

\(312-x=308,7\)

\(x=312-308,7\)

`x=3,3`

a: =>2x=-3/5-5/2=-6/10-25/10=-31/10

=>x=-31/20

b: =>1/2:x=-2/3+5/6=5/6-4/6=1/6

=>x=1/2:1/6=3

c: =>312-x=308,7

=>x=3,3

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

Bài 5:

a: x(x-4)=0

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

b: Đề thiếu vế phải rồi bạn

Bài 6:

a: \(\left(-5\right)\cdot\left(-6\right)\cdot\left(-4\right)\cdot2\)

\(=-\left(2\cdot5\right)\cdot\left(4\cdot6\right)\)

\(=-24\cdot10=-240\)

b: \(\left(-3\right)\cdot2\cdot\left(-8\right)\cdot5\)

\(=3\cdot2\cdot8\cdot5\)

\(=\left(3\cdot8\right)\cdot\left(2\cdot5\right)\)

\(=24\cdot10=240\)

`3(x-1)(x-5) =0`

`<=> (x-1) =0` hoặc `x-5 = 0`.

`<=> x =1` hoặc `x = 5`.

Vậy `x = 1` hoặc `x = 5.`

`b, 3x^2 + 7x = 10`.

`<=> 3x^2 + 7x - 10 = 0`

`<=> (3x+10)(x-1) =0`

`<=> 3x + 10 = 0` hoặc `x - 1=0`

`<=> x = -10/3` hoặc `x = 1.`

Vậy `x = -10/3` hoặc `x = 1.`

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-5x-15=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)