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29 tháng 6 2016

a)(x-1)(x2+x+1)-x(x+2)(x-2)=5

=>x3-1-4x-x3=5

=>x3-x3+4x-1=5

=>4x-1=5

=>4x=6

=>x=3/2

b)(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=15

=>x3-6x2+12x-8-x3+27+6x2+12x+6=15

=>(x3-x3)-(-6x2+6x2)+(12x+12x)-8+27+6=15

=>24x+25=15

=>24x=-10

=>x=-5/12

c)6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1

=>6x2+12x+6-2x3-6x2-6x-2+2x3-2=1

=>(6x2-6x2)+(12x-6x)-(-2x3+2x3)+6-2-2=1

=>6x+2=1

=>6x=-1

=>x=-1/6

29 tháng 9 2019

\(\frac{x}{3}+\frac{x^2}{2}=0\)

\(\Leftrightarrow\frac{2x+3x^2}{6}=0\Leftrightarrow3x^2+2x=0\)

\(\Leftrightarrow x\left(3x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)

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

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

\(\Leftrightarrow\left(x^2+4\right)\left(x+1\right)=0\)

Mà \(x^2+4>0\)nên \(x+1=0\Leftrightarrow x=-1\)

11 tháng 7 2017

\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow9x^3-18x^2+9x-29=0\)

\(\Leftrightarrow x=2,208024627\)

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

22 tháng 10 2015

5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=5

5x(x-3)^2-5(x-1)^3+15(x^2-2^2)=5

5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15x^2-60=5

5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5

30x-55=5

30x=60

x=2

15 tháng 7 2019

(x - 3).(x2 + 3x + 9) + x.(x + 2).(2 - x) = 1

<=> x3 - 27 + 4x - x3 = 1

<=> 4x - 27 = 1

<=> 4x = 1 + 27

<=> 4x = 28

<=> x = 7

=> x = 7

a: \(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}+\dfrac{4x^2}{x^2-9}\right):\dfrac{2x+1-x-3}{x+3}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9+4x^2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-2}\)

\(=\dfrac{4x^2-12x}{x-3}\cdot\dfrac{1}{x-2}=\dfrac{4x}{x-2}\)

b: \(2x^2-5x+2=0\)

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

=>x=1/2

Thay x=1/2 vào P, ta được:

\(P=\left(4\cdot\dfrac{1}{2}\right):\left(\dfrac{1}{2}-2\right)=2:\dfrac{-3}{2}=\dfrac{-4}{3}\)