a, \(x\) + 99: 3 = 55

    \(x\) + 33 = 55

     \(x\)        = 55 - 33

      \(x\)       = 22

b, (\(x\) - 25):15 = 20

       \(x\) - 25  = 20 x 15

       \(x\) - 25  = 300

        \(x\)         = 300 + 25

          \(x\)        = 325

c, (3\(x\) - 15).7 = 42

      3\(x\) - 15 = 42:7

       3\(x\) - 15 = 6

        3\(x\)         = 6 + 15

         3\(x\)         = 21

            \(x\)         = 21: 3

             \(x\)         = 7

d, (8\(x\) - 16).(\(x\) -5) = 0

       \(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=16:8\\x=5\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) {2; 5}

1: \(x^2+3x+2\)

\(=x^2+x+2x+2\)

=x(x+1)+2(x+1)

=(x+1)(x+2)

2: \(x^2+4x+3\)

\(=x^2+x+3x+3\)

=x(x+1)+3(x+1)

=(x+1)(x+3)

3: \(x^2+5x+4\)

\(=x^2+x+4x+4\)

=x(x+1)+4(x+1)

=(x+1)(x+4)

4: \(x^2-4x+3\)

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

=x(x-1)-3(x-1)

=(x-1)(x-3)

5: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

6: \(x^2-5x+4\)

\(=x^2-x-4x+4\)

=x(x-1)-4(x-1)

=(x-1)(x-4)

7: \(x^2-5x+6\)

\(=x^2-2x-3x+6\)

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

=(x-2)(x-3)

8: \(x^2+6x+5\)

\(=x^2+x+5x+5\)

=x(x+1)+5(x+1)

=(x+1)(x+5)

9: \(x^2-7x+10\)

\(=x^2-2x-5x+10\)

=x(x-2)-5(x-2)

=(x-2)(x-5)

10: \(x^2+8x+12\)

\(=x^2+2x+6x+12\)

=x(x+2)+6(x+2)

=(x+2)(x+6)

11: \(x^2-8x+16=x^2-2\cdot x\cdot4+4^2=\left(x-4\right)^2\)

12: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

13: \(x^2-8x+7\)

\(=x^2-x-7x+7\)

=x(x-1)-7(x-1)

=(x-1)(x-7)

14: \(x^2+9x+8\)

\(=x^2+x+8x+8\)

=x(x+1)+8(x+1)

=(x+1)(x+8)

15: \(x^2-9x+14\)

\(=x^2-2x-7x+14\)

=x(x-2)-7(x-2)

=(x-2)(x-7)

16: \(x^2+9x+18\)

\(=x^2+3x+6x+18\)

=x(x+3)+6(x+3)

=(x+3)(x+6)

17: \(x^2-9x+20\)

\(=x^2-4x-5x+20\)

=x(x-4)-5(x-4)

=(x-4)(x-5)

18: \(2x^2-3x+1\)

\(=2x^2-2x-x+1\)

=2x(x-1)-(x-1)

=(x-1)(2x-1)

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

2. \(x^2+4x+3=\left(x+1\right)\left(x+3\right)\)

3. \(x^2+5x+4=\left(x+1\right)\left(x+4\right)\)

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

5. \(x^2-4x+4=\left(x-2\right)^2\)

6. \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

7. \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

8. \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)

9. \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)

10. \(x^2+8x+12=\left(x+2\right)\left(x+6\right)\)

11. \(x^2-8x+16=\left(x-4\right)^2\)

12. \(x^2+8x+15=\left(x+3\right)\left(x+5\right)\)

13. \(x^2-8x+7=\left(x-1\right)\left(x-7\right)\)

14. \(x^2+9x+8=\left(x+1\right)\left(x+8\right)\)

15. \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)

16. \(x^2+9x+18=\left(x+3\right)\left(x+6\right)\)

17. \(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(18.2x^2-3x+1=2x^2-x-2x+1\)

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

Sửa lại câu d) là `25y^2`

`a)x^3-1`

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

`b)8x^3-y^3`

`=(2x)^3-y^3`

`=(2x-y)(4x^2+2xy+y^2)`

`c)x^2-8x+16`

`=x^2-2.x.4+4^2`

`=(x-4)^2`

`d)25y^2-1`

`=(5y)^2-1`

`=(5y-1)(5y+1(`

`e)27-8y^3`

`=3^3-(2y)^3`

`=(3-2y)(9+6y+4y^2)`

`f)2x^2-8x+8`

`=2(x^2-4x+4)`

`=2(x-2)^2`

a) x³ - 1 = x³ - 1³

              = (x - 1)(x² - x + 1)

b) 8x³ - y³ = (2x)³ - y³

                 = (2x - y)(4x² + 2xy + y²)

c) x² - 8x + 16 = x² - 2.4x + 4²

                        = (x - 4)²

d) đề có bị sai không , nên mình sưa lại đề nhé :

       25y² - 1 = (5y)² - 1²

                     = (5y - 1)(5y + 1)

e) 27 - 8y³ = 3³ - (2y)³

                  = (3 - 2y)(9 + 6y + 4y²)

f) 2x² - 8x + 8 = 2(x² - 4x + 4)

                       = 2(x - 2)²

 Chúc bạn học tốt

1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

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

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)