a) 3(x - 1) - 1 = 26

=> 3x - 3 = 26 + 1

=> 3x - 3 = 27

=> 3x = 27 + 3

=> 3x = 30

=> x = 30 : 3

=> x = 10

b) |x + 4| - 9 = (-2)³

=> |x + 4| - 9 = -8

=> |x + 4| = -8 + 9

=> |x + 4| = 1

=> \(\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\) 

=> \(\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

Vậy ...

c) 2x - 5 \(⋮\)x - 1

<=> 2(x - 1) - 3 \(⋮\)x - 1

<=> 3 \(⋮\)x - 1

<=> x - 1 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy ...

a) 3×(x-1) - 1 =26

    => 3×(x-1) = 26+1

    => 3×(x-1) = 27

    => x-1 = 27:3

    => x-1 = 9

    => x= 9+1 

    => x= 10

b) |x+4| - 9 = (-2)^3

   = | x+4| - 9 = -8

   = | x+4| = (-8)+9

   = | x+4| =1

  => x+4= 1 hay x+4= (-1)

Khi x+4= 1

      => x= 1-4

      => x= -3

Khi x+4= (-1)

      => x= (-1)+4

      => x= -5

Vậy: x= -3 hoặc x= -5

a, x= 3

b, ko có x thỏa mãn

c, x= 3

d, x= 2

a: x=3

b: \(2x-1=2\)

hay \(x=\dfrac{3}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

b: =>12-x=-7

hay x=19

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

Sos

hahaha

chịu

Chịu 

\(a,\left(-5\right).\left|x\right|=-75\)

\(\left|x\right|=\frac{-75}{-5}=15\)

\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

Vậy....

\(b,\left(-6\right)^3.x^2=-1944\)

\(-216.x^2=-1944\)

\(x^2=9\)

\(\Rightarrow x=\pm3\)

Vậy....

\(d,\left|9-x\right|=-7+64\)

\(\left|9-x\right|=57\)

\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)

Vậy...

\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)

\(\left|x+101\right|+16=215\)

\(\left|x+101\right|=199\)

\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)

Vậy..

hok tốt!!

a,\(\left(-5\right).\left|x\right|=-75\)

\(=>\left|x\right|=-75:\left(-5\right)=15\)

\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

b,\(\left(-6\right)^3.x^2=-1944\)

\(=>\frac{1944}{216}=x^2\)

\(=>x=\sqrt{\frac{1944}{216}}=3\)

CẢM ƠN MỌI NGƯỜI NHIỀU Ạ 

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )³ - x( x - 1 )( x + 1 ) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

a) (x+3)^2-4=0

=>(x+3)^2 = 4

=>(x+3)^2 = 2^2 = (-2)^2

=>x+3 = 2 hoặc -2

=> x= -1 hoặc -5