a) x + 1/15 = 1/3 - 1/4 = 4/12 - 3/12 = 1/12.

=> x = 1/12 - 1/15 = 5/60 - 4/60 = 1/60.

b) x - 5/12 = 1/3 + 1/4 = 4/12 + 3/12 = 7/12.

=> x = 7/12 + 5/12 = 1

c) x x 2/7 = 3/4 x 4/15 = 1/5

=> x = 1/5 : 2/7 = 1/5 x 7/2 = 7/10

d) x : 3/8 = 4/3 : 3/6 = 4/3 x 2 = 8/3

=> x = 8/3 x 3/8 = 1

4 - x =15 - 6

4 - x = 9

     x= 4-9

     x=-5

25-x=-1--30

25-x=29

    x=25-29

    x=-4

ai tk minh minh tk lai

A/  \(15-\left(4-X\right)=6\)

\(15-4+X=6\)

\(11+X=6\)

\(X=6-11\)

\(X=-5\)

B/   \(-30+\left(25-X\right)\)\(=-1\)

\(-30+25-X=-1\)

\(-5+X=-1\)

\(X=-1+5\)

\(X=4\)

a)15-(4-x)=6

     4-x=15-6

     4-x=9

       x=4-9

        x=-5

b)-30+(25-x)=-1

         25-x=-1+30

         25-x=29

              x=25-29

              x=-4 

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 \(x=-\dfrac{13}{24}\)

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:

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

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

Bài 2:
a) x + 5,7 = 18,6 - 10,3
    x + 5,7 = 8,3
    x          = 8,3 - 5,7 
    x          = 3,6
b) 6,4 . x  = 5 . 3,2
    6,4 . x  = 16
    6,4 . x  = 16 : 6,4 
    6,4 . x  = 2,5

B1

a) 3/5 . 20/18 : 2/9 .1/15

= 3/5 . 20/18 . 9/2 .1/15

= (3/5 . 1/15) . (20/18 . 9/2)

= 1/25 . 5

= 1/5

b. (5/2 + 1/8) : (1 - 7/16)

= 21/8 : 9/16

= 21/8 . 16/9

= 14/3 

B2:

\(a.x+5,7=18,6-10,3\\ x=18,6-10,3-5,7\\ x=18,6-\left(10,3+5,7\right)\\ x=18,6-16\\ x=2,6\\ b.6,4\cdot x=5\cdot3,2\\ \left(3,2\cdot2\right)\cdot x=5\cdot3,2\\ x=\dfrac{5}{2}\cdot\left(3,2:3,2\right)\\ x=\dfrac{5}{2}\)

c) x + 2/4 = 15/9 + 3/36

x + 2/4 = 7/4

x = 7/4 - 2/4

x = 5/4

d) x . 4/3 = 15/3 - 22/6

 x . 4/3 = 4/3 

x = 4/3 :4/3

 x = 1

B4:

Gọi số đầu tiên là a

Vì tổng của 4 số tự nhiên liên tiếp  = 2010

=> a + (a+1) + (a+2) + (a+3) =2010

=> a4 + 6 = 2010

=> a4 = 2004

=> a = 501

Số thứ 2 là:

501 + 1 = 502

Số thứ 3 là:

502 + 1 = 503

Số thứ 4 là :

503 + 1 = 504

a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)

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

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

\(x=-\dfrac{7}{10}\)

b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)

\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)

\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)

\(x=\dfrac{60}{13}:\dfrac{12}{13}\)

\(x=5\)

a) \(15-\left(4-x\right)=6\)

\(\Rightarrow4-x=15-6\)

\(\Rightarrow4-x=9\)

\(\Rightarrow x=4-9\)

\(\Rightarrow x=-5\)

b)\(-30+\left(25-x\right)=-1\)

\(\Rightarrow25-x=-1+30\)

\(\Rightarrow25-x=29\)

\(\Rightarrow x=25-29\)

\(\Rightarrow x=-4\)

a, \(x\) + 99: 3 = 55

    \(x\) + 33 = 55

     \(x\)        = 55 - 33

      \(x\)       = 22

b, (\(x\) - 25):15 = 20

       \(x\) - 25  = 20 x 15

       \(x\) - 25  = 300

        \(x\)         = 300 + 25

          \(x\)        = 325

c, (3\(x\) - 15).7 = 42

      3\(x\) - 15 = 42:7

       3\(x\) - 15 = 6

        3\(x\)         = 6 + 15

         3\(x\)         = 21

            \(x\)         = 21: 3

             \(x\)         = 7

d, (8\(x\) - 16).(\(x\) -5) = 0

       \(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=16:8\\x=5\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy \(x\) \(\in\) {2; 5}