2/3 + 3/4=17/12
9/4 + 3/5=57/20
5/24 + 1/4=11/24
3/15 - 5/35=2/35        18/27 - 2/6=1/3        37/12 - 3=1/12
11/9 x 3/22=1/6         7/13 x 13/7=1       4 x6/7=24/7
2/5 : 3/10=4/3           3/8 : 9/4=1/6            8/21 : 4/7=2/3

2/3 + 3/4

= 17/12
9/4 + 3/5

= 57/20
5/24 + 1/4

=11/24
3/15 - 5/35   

=2/35   

18/27 - 2/6   

=1/3   

37/12 - 3

= 1/12
11/9 x 3/22 

=1/6       

7/13 x 13/7 

=1   

 4 x6/7

=24/7
2/5 : 3/10

=4/3         

 3/8 : 9/4   

=1/6       

8/21 : 4/7

= 2/3

1=17/12
2=57/20
3=11/24
4=2/35        7=1/3        10=1/12
5=1/6         8=1       11=24/7

6=4/3           9=1/6            12=2/3

1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2

2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2

7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7

3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0

8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9

4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2

9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9

5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8

10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=\left(-2x+7\right)\cdot1\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

\(=-10x-50\)

mn. giúp em nha 

\(a.\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow x=\dfrac{18\cdot5}{6}=15\\ \text{Vậy}\text{ }x=15.\)

\(b.\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow x=\dfrac{-21\cdot4}{3}=28\\ \text{ }\text{ }\text{ }\text{ }\text{Vậy }x=28.\)

\(c.\dfrac{x}{4}=\dfrac{21}{28}\Rightarrow x=\dfrac{21\cdot4}{28}=3\\ \text{Vậy }x=3.\)

\(d.\dfrac{-8}{2x}=\dfrac{3}{-9}\Rightarrow x=\dfrac{-8\cdot\left(-9\right)}{3}:2=12\\ \text{Vậy }x=12.\)

\(e.\dfrac{-4}{11}=\dfrac{x}{22}=\dfrac{40}{z}\\ \Rightarrow x=\dfrac{-4\cdot22}{11}=-8\\ \Rightarrow z=\dfrac{22\cdot40}{-8}=-110\\ \text{Vậy }x=-8;z=-110.\)

\(f.\dfrac{-3}{4}=\dfrac{x}{20}=\dfrac{21}{y}\\ \Rightarrow x=\dfrac{-3\cdot20}{4}=-15\\ \Rightarrow y=\dfrac{21\cdot20}{-15}=-28\\ \text{Vậy }x=-15;y=-28.\)

\(g.\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\\ \Rightarrow x=\dfrac{-4\cdot\left(-10\right)}{8}=5\\ \Rightarrow y=\dfrac{-7\cdot\left(-10\right)}{5}=14\\ \Rightarrow z=\dfrac{-7\cdot\left(-24\right)}{14}=12\\ \text{Vậy }x=5;y=14;z=12.\)

\(h.\dfrac{x}{4}=\dfrac{9}{x}\\ \Rightarrow x\cdot x=9\cdot4\\ \Rightarrow x\cdot x=36\\ \Rightarrow x\cdot x=6\cdot6\\ \text{Vậy }\text{cả hai }x=6.\)

nhiều quá :((

\(a,2\left(x-5\right)-3\left(x+7\right)=14\)

\(2x-10-3x-21=14\)

\(-x-31=14\)

\(-x=45\)

\(x=45\)

\(b,5\left(x-6\right)-2\left(x+3\right)=12\)

\(5x-30-2x-6=12\)

\(3x-36==12\)

\(3x=48\)

\(x=16\)

\(c,3\left(x-4\right)-\left(8-x\right)=12\)

\(3x-12-8+x=0\)

\(4x-20=0\)

\(4x=20\)

\(x=5\)

Cố nốt nha bn ! 

cảm ơn, bn nha:)))

mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???

d, \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)

\(\Leftrightarrow-21x+35+14x-28=28\Leftrightarrow-7x=21\Leftrightarrow x=-3\)

e, \(5\left(3-2x\right)+5\left(x-4\right)=6-4x\)

\(\Leftrightarrow15-6x+5x-20=6-4x\Leftrightarrow-5-x=6-4x\)

\(\Leftrightarrow-11+3x=0\Leftrightarrow x=\frac{11}{3}\)

f, \(-5\left(2-x\right)+4\left(x-3\right)=10x-15\)

\(\Leftrightarrow-10+5x+4x-12=10x-15\Leftrightarrow-22+9x=10x-15\)

\(\Leftrightarrow-7-x=0\Leftrightarrow x=-7\)

a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)

   \(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

    \(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

     \(\dfrac{72}{17x}\)        = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)

      \(\dfrac{72}{17x}\)      = - \(\dfrac{1}{14}\)

      17\(x\)       = 72.(-14)

       17\(x\)     = - 1008

         \(x\)       = - 1008 : 17

         \(x\)       = - \(\dfrac{1008}{17}\)

Vậy \(x\) \(=-\dfrac{1008}{17}\)

b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))³ = - \(\dfrac{24}{27}\)

          - \(\dfrac{32}{27}\)  + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))³ 

          (3\(x-\dfrac{7}{9}\))³ = - \(\dfrac{8}{27}\)

         (3\(x-\dfrac{7}{9}\))³ = (- \(\dfrac{2}{3}\))³

           3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)

           3\(x\)        = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
           3\(x\)        = \(\dfrac{1}{9}\)

             \(x\)        = \(\dfrac{1}{9}\) : 3

             \(x\)       = \(\dfrac{1}{27}\)

Vậy \(x=\dfrac{1}{27}\)

1/ -7264 + (1543 + 7264)
=-7264 + 1543 + 7264=1543
2/ (144 – 97) – 144

=144-97-144=-97

3/ (-145) – (18 – 145)(Vì có dấu trừ ở trước ngoặc nên p đổi dấu)

=-145-18+145=-18

4/ 111 + (-11 + 27)

=111-11+27=137

1/ -7264 + (1543 + 7264)
=-7264 + 1543 + 7264=1543
2/ (144 – 97) – 144
=144-97-144=-97
3/ (-145) – (18 – 145)(Vì có dấu trừ ở trước ngoặc nên p đổi dấu)
=-145-18+145=-18
4/ 111 + (-11 + 27)
=111-11+27=137

Sao giống y chang chương trình của mình vậy :)