a) Ta có: \(\left(2x-5\right)^3=216\)

\(\Leftrightarrow2x-5=6\)

\(\Leftrightarrow2x=11\)

hay \(x=\dfrac{11}{2}\)

b) Ta có: \(2x-3⋮x+4\)

\(\Leftrightarrow-11⋮x+4\)

\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-3;-5;7;-15\right\}\)

Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da

-2x - 40 = (5 - x) - (-15 + 60)

-2x - 40 = (5 - x) - 45

-2x - 40 = -40 - x

-2x - x = 40 - 40

      -3x = 0

         x = 0

2(x - 4) - 3(x + 7) = 14

    2x - 8 - 3x - 21 = 14

                 -x - 29 = 14

                        -x = 14  + 29

                        -x = 43 

                         x = -43

-7(5 - x) - 2(x - 10) = 15

-35 + 7x - 2x + 20 = 15

              -15 + 5x = 15

                       5x = 15 + 15

                       5x = 30

                         x = 6

         \(-2x-40=\left(5-x\right)-\left(-15+60\right)\)

\(\Leftrightarrow\)\(-2x-40=5-x+15-60\)

\(\Leftrightarrow\)\(-2x-40=-x-40\)

\(\Leftrightarrow\)\(-2x+x=40-40\)

\(\Leftrightarrow\)\(x=0\)

Vậy....

\(x+18=2-2x\)

\(2x+x=2-18\)

\(3x=-20\)

\(x=-20:3\)

\(x=\frac{-20}{3}\)

a) x - 14 = 3x + 18

=> x - 3x = 18 + 14

=> -2x = 32

=> x = 32 : (-2)

=> x = -16

b) 2(x - 5) - 3(x - 4) = -6 + 15.(-3)

=> 2x - 10 - 3x + 12 = -6 - 45

=> -x + 2 = -51

=> -x = -51 - 2

=> -x = -53

=> x = 53

c) (x + 7)(x - 9) = 0

=> \(\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=9\end{cases}}\)

Vậy ...

d) |2x - 5| - 7 = 22

=> |2x - 5| = 22 + 7

=> |2x - 5| = 29

=> \(\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\)

=> \(\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\)

=> \(\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)

Vậy ...

\(x-14=3x+18\)

\(x-3x=14+18\)

\(-2x=32\)

\(-x=16\)

\(\Leftrightarrow x=-16\)

d) \(\left|2x-5\right|-7=22\)

\(\Rightarrow\left|2x-5\right|=22+7\)

\(\Rightarrow\left|2x-5\right|=29\)

\(\Rightarrow\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)

Vậy x = 17 hoặc x = -12

Ở câu b, bậc của y là bậc nhất nên có thể rút y theo x

\(y=\frac{112-2x^2+x}{2x+1}=\frac{-x\left(2x+1\right)+2x+1+111}{2x+1}=-x+1+\frac{111}{2x+1}\)

\(\Rightarrow2x+1\in\text{Ư}\left(111\right)=\left\{111;37;3;1;-111;-37;-3;-1\right\}\)

\(\Rightarrow x\in\left\{...\right\}\)

a) \(x+15=18-\left(-27\right)\)

\(\Leftrightarrow x+15=18+27\)

\(\Leftrightarrow x+15=45\)

\(\Leftrightarrow x=45-15\)

\(\Leftrightarrow x=30\)

b) \(-x+7=28\)

\(\Leftrightarrow-x=28-7\)

\(\Leftrightarrow-x=21\)

\(\Leftrightarrow x=-21\)

c) \(20-\left(-x\right)=14\)

\(\Leftrightarrow20+x=14\)

\(\Leftrightarrow x=14-20\)

\(\Leftrightarrow x=-6\)

d) \(2x+5=x-14\)

\(\Leftrightarrow2x+5-\left(x-14\right)=0\)

\(\Leftrightarrow2x+5-x+14=0\)

\(\Leftrightarrow x+19=0\)

\(\Leftrightarrow x=0-19\)

\(\Leftrightarrow x=-19\)

e) \(x+17=-8-20\)

\(\Leftrightarrow x+17=-\left(8+20\right)\)

\(\Leftrightarrow x+17=-28\)

\(\Leftrightarrow x=-28-17\)

\(\Leftrightarrow x=-\left(28+17\right)\)

\(\Leftrightarrow x=-45\)

f) \(3x+15=2x+6\)

\(\Leftrightarrow3x+15-\left(2x+6\right)=0\)

\(\Leftrightarrow3x+15-2x-6=0\)

\(\Leftrightarrow x+9=0\)

\(\Leftrightarrow x=0-9\)

\(\Leftrightarrow x=-9\)

2,

a.\(\left|a+3\right|=\left|-5\right|\)

\(a+3=5\)

\(a=2\)

Vậy a=2

b.\(\left|a-1\right|=2+\left|-3\right|\)

\(a-1=2+3\)

\(a-1=5\)

\(a=6\)

Vậy a=6

5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39

<=> 5x - 5 - 7x + 14 = 2x - 39

<=> 5x - 7x - 2x = -39 + 5 - 14

<=> -4x = -48

<=> x = 12

x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)

\(\Rightarrow\chi-3-28+3=-3\chi-3\)

\(\Rightarrow\chi-28=11\chi\)

\(\Rightarrow\chi-11\chi=28\)

\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\) 

HTDT-sai-rùi

x-3-14(x-2)=-3x-3

x-3+28-14x=-3x-3

-13x+35=-3x-3

-13x+3x=-3-35

-10x=-38

x=38/10=19/5