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

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

__________________________________________

`b)(x-3)^3+3-x=0` nhỉ?

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

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

__________________________________________

`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

__________________________________________

`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

__________________________________________

`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

\(\Leftrightarrow x^2-6x+9-x^2+4=1\)

=>-6x=-12

hay x=2

Nhanh giúp mình với mn

Tìm x hả?

8/x+1/5=2/5+1/3.

8/x+1/5=11/15

x=11/15-1/5

x=8/15

\(\frac{1}{3}\times X=\frac{1}{6}\)

\(X=\frac{1}{6}\div\frac{1}{3}\)

\(X=\frac{1}{2}\)

lấy 1 phần 6 : 1 phần 3

bằng 1/2 nhé bạn

= x³ + 3³ -x(x² -1) -27 =0 ( tổng các lập phuong)

x =0 

CX100%

bạn chỉ cần phá hết hằng đẳng thức ra thôi 

câu 1

(x+3)^2+(x-2)(x+2)-2(x-1)^2=7

<=> x^2+6x+9+x^2-4-2x^2-4x-2-7=0

<=> x^2+x^2-2x^2+6x-4x+9-4-2-7=0

<=> 2x-4=0

<=> x=2

câu 2

(x+3)(x^2-3x+9)-x(x-1)(x+1)-27=0

<=> x^3+27-x(x^2-1)-27=0

<=> x^3-x^3+x+27-27=0

<=>x=0

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

(x+1/10).5/6=3-7/6

(x+1/10).5/6=18/6-7/6

(x+1/10).5/6=11/6

x+1/10=11/6:5/6

x+1/10=11/6.6/5

x+1/10=11/5

x        =11/5-1/10

x=22/10-1/10

x=21/10

dấu . là dấu nhân nha

Chúc bạn học tốt

Bạn ơi k đúng để làm j vậy kkk

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)

Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)

x = -2014

ti-ck nha

.........

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\left(x+1\right)=0\)

\(\Rightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)

\(x=-1\)

Vậy \(x=-1\)

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

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

\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)

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

\(=x^3+3x^2+3x+1-x^3+1-2=0\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow3x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)