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2 tháng 11 2017

a) (x + 2)(x + 3) - (x - 2)(x + 5) = 6

x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 6

x2 + 5x + 6 - x2 - 3x + 10 = 6

2x +16 = 6

\(\Rightarrow\) 2x = -10

\(\Rightarrow\) x = -5

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

6x2 + 27x + 4x + 18 - (6x2 + x + 12x + 2) = x + 1 - x + 6

6x2 + 31x + 18 - 6x2 - 13x - 2 = 7

18x + 16 = 7

\(\Rightarrow\) 18x = -9

\(\Rightarrow\) x = -0.5

c) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0

3(6x2 - 2x - 3x + 1) - (18x2 - 2x - 27x + 3) = 0

3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0

18x2 - 15x + 3 - 18x2 + 29x - 3 = 0

14x = 0

\(\Rightarrow\) x = 0

2 tháng 11 2017

Thank you very much!

16 tháng 12 2022

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

16 tháng 12 2022

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

22 tháng 12 2020

Rảnh rỗi thật sự .-.

undefined

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

25 tháng 7 2021

a) (x-2)3+6(x+1)2-x3+12=0

⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0

⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

24 tháng 8 2017

a) (3x+2)(2x+9) - (x+2)(6x+1) = (x+1) - (x-6)

<=> 6x2 + 27x + 4x + 18 - 6x2 - x - 12x - 2 = x+1 - x+6

<=> 18x + 16 = 7

<=> 18x = -9

<=> x = \(-\dfrac{1}{2}\)

b) 3(2x-1)(3x-1) - (2x-3)(9x-1) = 0

<=> 3.(6x2-2x-3x+1) - (18x2-2x-27x+3) = 0

<=> 3.(6x2-5x+1) - 18x2+29x-3 = 0

<=> 18x2-15x+3 - 18x2+29x - 3 = 0

<=> 14x = 0

<=> x = 0

4 tháng 7 2017

\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x^2-x-6x-6\)

\(\Leftrightarrow18x+16=x^2-7x-6\)

\(\Leftrightarrow x^2-7x-18x=16+6\)

\(\Leftrightarrow x^2-15x=22\)

\(\Leftrightarrow x^2-15x-22=0\)

......

11 tháng 6 2018

\(\Leftrightarrow\left(6x^2+27x+4x+18\right)-\left(6x^2+x+12x+2\right)=x-1-x+6\)

\(\Leftrightarrow6x^2+31x+18-6x^2-x-12x-2=7\)

\(\Leftrightarrow18x+16=7\)

\(\Leftrightarrow18x=-9\)

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

11 tháng 6 2018

                       \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)

\(3x\left(2x+9\right)+2\left(2x+9\right)-x\left(6x+1\right)-2\left(6x+1\right)=x-1-x+6\)

                 \(6x^2+27x+4x+18-6x^2-x-12x-2=5\)

        \(6x^2+\left(27x+4x\right)+18-6x^2-\left(12x+x\right)-2=5\)

                                 \(6x^2+31x+18-6x^2-13x-2=5\)

                    \(\left(6x^2-6x^2\right)+\left(31x-13x\right)+\left(18-2\right)=5\)

                                                                           \(18x+16=5\)

                                                                                     \(18x=5+16\)

                                                                                     \(18x=21\)

                                                                                          \(x=21:18\)

                                                                                          \(x=\frac{7}{6}\)

                                                Vậy \(x=\frac{7}{6}\)

P/s: Mình mới lớp 6 nên hi vọng bn xem bài của mik thật kĩ xem có sai sót không,cảm ơn.

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

\(\Leftrightarrow\left(x^3-8\right)\left(x^3+8\right)-x^6+2x-1=0\)

\(\Leftrightarrow x^6-64-x^6+2x-1=0\)

\(\Leftrightarrow2x-65=0\)

\(\Leftrightarrow2x=65\)

hay \(x=\frac{65}{2}\)

Vậy: \(x=\frac{65}{2}\)

c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27-x\left(x+2\right)\left(x-2\right)-1=0\)

\(\Leftrightarrow x^3-27-x\left(x^2-4\right)-1=0\)

\(\Leftrightarrow x^3-27-x^3+4x-1=0\)

\(\Leftrightarrow4x-28=0\)

\(\Leftrightarrow4x=28\)

hay x=7

Vậy: x=7

11 tháng 1 2023

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

 

11 tháng 1 2023

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

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

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

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

`<=>[(x=-5),(x=1):}`