giúp mình nha mn

giúp mk với

Đặt a = 2x - 5; b = 3x - 4 => x + 1 = b - a

Khi đó ta có pt: a³ - b³ + (b - a)³ = 0

<=> (b - a)³ - (b³ - a³) = 0

<=> b³ - a³ - b³ + a³ - 3ab(b - a)= 0

<=> 3ab(b - a) = 0 <=> a = 0 hoặc b = 0 hoặc a= b

* Với a = 0, ta có: 2x - 5 = 0 <=> x = \(\frac{5}{2}\)

* Với b = 0, ta có: 3x - 4 = 0 <=> x = \(\frac{4}{3}\)

* Với a = b, ta có: 2x - 5 = 3x - 4 <=> x = -1

Vậy S = {-1;\(\frac{4}{3}\);\(\frac{5}{2}\)}

D

(x² -2x+1)-4=0

= (x-1)²-4=0

=> (x-1)²=4

=> x=3

2 - x 2 + 3 x < 2 2

⇔ − x 2  + 3x < 2

⇔  x 2  − 3x + 2 > 0

\(\dfrac{x+3}{x+2}+\dfrac{x}{2-x}=\dfrac{5x}{x^2-4}\)

\(\Leftrightarrow\dfrac{x+3}{x+2}-\dfrac{x}{x-2}=\dfrac{5x}{\left(x-2\right)\left(x+2\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+2\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne2\end{matrix}\right.\)

Ta có : \(\dfrac{x+3}{x+2}-\dfrac{x}{x-2}=\dfrac{5x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{\left(x-2\right)\left(x+2\right)}\)

`=> x^2 -2x +3x-6 - x^2 -2x -5x=0`

`<=>-6x -6=0`

`<=>-6x=6`

`<=>x=-1(t/m)`

=>(x+3)(x-2)-x(x+2)=5x

=>x^2+x-6-x^2-2x=5x

=>5x=-x-6

=>6x=-6

=>x=-1

1a.4,9,16,25,36,49,64,91,110,121,144,169,400,900,1600

b. 8,27,64,125,1000

C1111^2=1234321

2. x=6   ,x=3,x=0, x k thuộc n, x k thuộc n

sao bai 2 ban lamminh ko hieu ?