2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

b) (x – 4). (x2 + 4x + 16) – x. (x2 - 6) = 2

⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – (x3 - 6x ) – 2= 0

⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – x3 + 6x – 2= 0

⇔ 6x – 66 =0

⇔ 6x = 66

⇔ x = 66 : 6

⇔ x = 11

Vậy x = 11

a) x^3 - 64 - x^3 +6x = 2

(x^3 - x^3) + 6x = 2+64 quy tắc chuyển vế nhé bạn

6x = 66

x = 66:11

x = 6

ta có

\(5x=-3y=4z\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)

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

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

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

\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)

Vậy: ...

\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy: ...

\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

Vậy: ...

\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy: ...

\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy: ...

\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)

Vậy: ...