a: -3/7=-27/63

5/9=35/63

b: -42/108=-7/12

5/12=5/12

\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\\ =\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\\ =\dfrac{\left(x+6\right)\left(x^2-2x+6\right)}{2x\left(x+6\right)}=\dfrac{x^2-2x+6}{2x}\)

\(\dfrac{x^2+2x}{2x+12}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+6\right)}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x\left(x^2+2x\right)+2\left(x+6\right)\left(x-6\right)+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+2x^2+2\left(x^2-36\right)+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+4x^2+36-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\)

\(=\dfrac{\left(x^3+6x^2\right)+\left(-2x^2-12x\right)+\left(6x+36\right)}{2x\left(x+6\right)}\)

\(=\dfrac{x^2\left(x+6\right)-2x\left(x+6\right)+6\left(x+6\right)}{2x\left(x+6\right)}\)

\(=\dfrac{\left(x+6\right)\left(x^2-2x+6\right)}{2x\left(x+6\right)}\)

\(=\dfrac{x^2-2x+6}{2x}\)

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)

\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(2A=3A-A\)

\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(2A=\frac{3}{4}-\frac{1}{927}\)

\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)

\(A=\frac{182}{243}:\frac{1}{2}\)

\(A=\frac{364}{243}\)

A=1/4+1/12+1/36+1/108+1/324+1/972

=243/972+81/972+27/972+9/972+3/972+1/972

=364/972

=91/243

\(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}\)

\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)

\(B=1.9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)

\(B=9-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\right]\)

\(B=9-\left(1-\frac{1}{56}\right)\)

\(B=9-\frac{55}{56}\)

\(B=\frac{449}{56}\)

a) 42 - x = 5

x = 42 - 5

x = 37

b) 15.(x + 4) = 75

x + 4 = 75 : 15

x + 4 = 5

x = 5 - 4

x = 1

d) x + 18 = 12

x = 12 - 18

x = -6

e) 2x - 15 = -19

2x = (-19) + 15

2x = -4

x = -4 : 2

x = -2

f) 36 : (x^3 - 12) = -3

x^3 - 12 = 36 : (-3)

x^3 - 12 = -12

x^3 = 0

=> x = 0

g)x + 15 = 35

x = 35 - 15

x = 20

h) 2x - 13 = 3^2

2x - 13 = 9

2x = 9 + 13

2x = 22

x = 22 : 2

x = 11

a) 42-x=5

         x=42-5

         x= 37

b)15*(x+4)=75

        x+4=75:15

        x+4=5

        x    =5-4

       x     = 1

a) 42 - x = 5

x = 42 - 5

x = 37

b) 15.(x + 4) = 75

x + 4 = 75 : 15

x + 4 = 5

x = 5 - 4

x = 1

d) x + 18 = 12

x = 12 - 18

x = -6

e) 2x - 15 = -19

2x = (-19) + 15

2x = -4

x = -4 : 2

x = -2

f) 36 : (x^3 - 12) = -3

x^3 - 12 = 36 : (-3)

x^3 - 12 = -12

x^3 = 0

=> x = 0

g)x + 15 = 35

x = 35 - 15

x = 20

h) 2x - 13 = 3^2

2x - 13 = 9

2x = 9 + 13

2x = 22

x = 22 : 2

x = 11

A = \(\dfrac{2}{3}\) + \(\dfrac{3}{18}\) + \(\dfrac{1}{42}\) + \(\dfrac{2}{63}\) + \(\dfrac{3}{108}\)

A = \(\dfrac{2}{1\times3}\) + \(\dfrac{3}{3\times6}\) + \(\dfrac{1}{6\times7}\)+ \(\dfrac{2}{7\times9}\) + \(\dfrac{3}{9\times12}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{12}\)

A = 1 - \(\dfrac{1}{12}\)

A = \(\dfrac{11}{12}\) 

a) \(⇔x^2-9x+20=12 \)

\(⇔x^2-9x+8=0\)

\(⇔x^2-x-8x+8=0\)

\(⇔(x-1)(x-8)=0\)

\(⇔\left[\begin{array}{} x-1=0\\ x-8=0 \end{array}\right.⇔\left[\begin{array}{} x=1\\ x=8 \end{array}\right.\)

b) \(⇔4x^2-12x+8=3\)

\(⇔4x^2-12x+5=0\)

\(⇔(2x-1)(2x-5)=0\)

\(⇔\left[\begin{array}{} 2x-1=0\\ 2x-5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{1}{2}\\ x=\frac{5}{2} \end{array}\right.\)

c) \(⇔x^2+x-30=42\)

\(⇔x^2+x-72=0\)

\(⇔(x-9)(x+8)=0\)

\(⇔\left[\begin{array}{} x-9=0\\ x+8=0 \end{array}\right.⇔\left[\begin{array}{} x=9\\ x=-8 \end{array}\right.\)

d) \(⇔2x^2+5x-3=-6\)

\(⇔2x^2+5x+3=0\)

\(⇔(x+1)(2x+3)=0\)

\(⇔\left[\begin{array}{} x+1=0\\ 2x+3=0 \end{array}\right.⇔\left[\begin{array}{} x=-1\\ x=-\frac{3}{2} \end{array}\right.\)

Đề bài là j vậy bạn?