a, \(\dfrac{x^2-49}{x-7}\) + x - 2 = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x - 2 = x + 7 + x - 2 = 2x + 5

b, \(\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right)\) . \(\dfrac{x^2+6x}{2x-6}\) 

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

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

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

= \(\dfrac{6}{x-6}\)

1. = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x-2

    = x+7 +x-2

    = 2x-5

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

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

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

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

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

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

     = \(\dfrac{6}{x-6}\)

Chúc bạn học tốt!

\(\left(5x\right)^2=\left(\dfrac{6}{7}\right)^2\)

⇒5x=6/7 =>x=6/35

5x=-6/7   =>x=-6/36

\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)

=>x-2/9=4/9

=>x=6/9

\(a)\dfrac{x^2}{6}=\dfrac{36}{x}\)

\(=>x^3=36.6\)

\(=>x^3=6^3\)

\(=>x=6\)

(câu b thiếu dữ kiện)

                  áp dụng dãy tỉ số bằng nhau ta có 

                      x/3=y/7=z/2=x+y+z/3+7+2=-16/12=-4/3

                   =>x/3=-4/3=>x=-4/3X3=-4

                  =>y/7=-4/3=>y=-4/3X7=-9,(3)

                 =>z/2=-4/3=>z=-4/3X2=-2(6)

b, \(\dfrac{x}{3}\)= \(\dfrac{y}{7}\)= \(\dfrac{z}{2}\)  Mà x+ y+ z= -16
             
    _xl mk bị thiếu mất vế sau ^^_

`(5/6 -x+7/12) : ( 11/24 -1/8)=11/36`

`=>(5/6 -x+7/12) : (11/24 - 3/24)=11/36`

`=>(5/6 -x+7/12) : 8=11/36`

`=>5/6 -x+7/12=11/36 xx 8`

`=>5/6 -x+7/12=22/9`

`=> x+7/12=5/6-22/9`

`=> x+7/12=-29/18`

`=>x=-29/18 -7/12`

`=>x=-79/36`

79/36 nhé

\(x:\left[\dfrac{8}{5}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{5}\right]=\dfrac{15}{7}+\dfrac{6}{5}\left[\left(2\dfrac{1}{7}\right)^2-\dfrac{50}{49}\right]\)

\(\Leftrightarrow x:\left[\dfrac{32}{45}-\dfrac{18}{45}\right]=\dfrac{15}{7}+\dfrac{6}{5}\cdot\left(\dfrac{225}{49}-\dfrac{50}{49}\right)\)

\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{15}{7}+\dfrac{6}{5}\cdot\dfrac{25}{7}\)

\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{45}{7}\)

\(\Leftrightarrow x=2\)

Lời giải:

$z=(x+y+z)-(x+y)=21-4=17$

$y=z-5=17-5=12$

$2k=z+x=(x+y+z)-y=21-12=9$

$k=\frac{9}{2}$

Không đáp án nào đúng.

a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)

\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)

\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)

\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)

\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)

Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)

Nên: \(x+100=0\)

\(x=-100\)

a: Sửa đề; \(A=\dfrac{7.2:2\cdot28.6+1.43\cdot2\cdot64}{1+3+5+7+...+49-339}\) 

\(=\dfrac{3.6\cdot28.6+2.86\cdot64}{1+3+5+...+49-339}\)

\(=\dfrac{2.86\left(64+36\right)}{25^2-339}=\dfrac{286}{286}=1\)

b: =>2(x+7/8)=6*13/4=78/4=39/2

=>x+7/8=39/4

=>x=71/8

a) \(\dfrac{3}{x-7}+\dfrac{2}{x+7}=\dfrac{5}{x^2-49}\)

(ĐKXĐ: x khác 7; x khác -7)

<=>\(\dfrac{3.\left(x+7\right)}{\left(x-7\right).\left(x+7\right)}+\dfrac{2.\left(x-7\right)}{\left(x+7\right).\left(x-7\right)}=\dfrac{5}{\left(x+7\right).\left(x-7\right)}\)

=> 3x + 21 + 2x - 14 = 5

<=> 3x + 2x = 5 + 14 - 21

<=> 5x = -2

<=> x = \(\dfrac{-2}{5}\)

Vậy S = { \(\dfrac{-2}{5}\) }

b) \(\dfrac{2x-1}{3}-\dfrac{x+3}{2}>1+\dfrac{5x}{6}\)

<=> \(\dfrac{2.\left(2x-1\right)}{3.2}-\dfrac{3.\left(x+3\right)}{3.2}>\dfrac{1.6}{6}+\dfrac{5x}{6}\)

=> 4x - 2 - 3x - 9 > 6 + 5x

<=> 4x - 3x - 5x > 6 + 9 + 2

<=> -4x > 17

<=> \(\dfrac{-17}{4}\)

Vậy S = { \(\dfrac{-17}{4}\) }