`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

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

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

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

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

a.\(\dfrac{1}{3}\) + x  = \(\dfrac{5}{6}\)

       x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)

      x = \(\dfrac{1}{2}\)

b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\) 

   | x-1|        = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)

  |x-1|        = \(\dfrac{3}{2}\)

\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1

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

             \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)

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

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

                    \(\dfrac{x}{2}\)    = -2

                     \(x\) = -4

d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)

(x+2)² = 27.3

(x+2) =9²

\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)

a) (x-1)(x-2)=0

x-1=0 --> x = 1

x-2=0 --> x = 2

d) x^2(x + 1) + 27(x + 1)=0

(x+1)(x^2+27)=0

x+1=0 --> x = -1

x^2+27=0 (vô lí)

a) \(x\left(x-2\right)-x+2=0\)

\(x\left(x-2\right)-\left(x-2\right)=0\)

\(\left(x-1\right)\left(x-2\right)=0\)

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

a) 

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

b) \(\left(3-2x\right)^2-\left(x-1\right)^2=0\)

\(\left(3-2x-x+1\right)\left(3-2x+x-1\right)=0\)

\(\left(3-3x+1\right)\left(3-x-1\right)=0\)

TH1:3-3x+1=0⇒x\(=\dfrac{4}{3}\)

TH2:3-x-1=0⇒x=2

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

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a) (3x-15)⁷ = 0

3x-15 = 0

3x = 0+15

3x = 15

x = 15:3

x = 5

b) 4^2x-6 = 1

 2x-6 = 0

2x = 0+6

2x = 6

x = 6:2

x = 3

c) Tớ ko bít 

d) (x - 6)³ = (x - 6)²

Th₁:

x - 6 = 1

x = 1 + 6

x = 7

Th₂:

x - 6 = 0

x = 6

Vậy x = 7

      x = 6

--thodagbun--

a, (3x-15)^7=0 <=> 3x-15=0 <=> x=5

b, 4²x+6=1 <=> 16x=-5 <=>x=-5/16

c, \(\dfrac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=1\)

TH1: 10x-20 = 0 <=> x=2

TH2: 3-x=1 <=> x=2

Vậy x=2

d, (x-6)^3 = (x-6)^2

<=> (x-6)^2.[(x-6)-1]=0

<=> (x-6)^2=0 hoặc (x-6)-1=0

<=> x=6 hoặc x=7

a: \(\Leftrightarrow x^2-2x-8-x^2=36\)

=>-2x=44

hay x=-22

b: \(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)

=>-34x=3

hay x=-3/34

c: =>(x-10)(x-1)=0

=>x=10 hoặc x=1

a: \(\Leftrightarrow\left(x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27.\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-27=0.\)

\(\Leftrightarrow x^3-x^3+4x^2-4x=0.\)

\(\Leftrightarrow4x\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0.\\x-1=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=1.\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}.\)

a) x + \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)

    x = \(\dfrac{5}{3}\) - \(\dfrac{3}{4}\)

    x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)  

    x =  \(\dfrac{11}{12}\) 

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

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

    x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\) 

    x = \(\dfrac{25}{6}\)

nhanh lên nhé 

sos