Ta có :

x²+2x-y²+2y = x² - y² + 2x+2y

= (x-y)(x+y) + 2(x+y) 

= (x-y+2)(x+y)

a, \(\dfrac{2-5x}{3}+\dfrac{2x^2-x}{2}>\dfrac{x\left(3x-1\right)}{3}-\dfrac{5x}{4}\)

\(\Rightarrow8-20x+12x^2-6x>4x\left(3x-1\right)-15x\)

\(\Leftrightarrow8-26x+12x^2>12x^2-19x\Leftrightarrow8>7x\Leftrightarrow x< \dfrac{8}{7}\)

b, \(\Rightarrow20x+10x+5>30x-2\Leftrightarrow5>-2\)(luôn đúng) 

Vậy bft luôn có nghiệm 

Ta có \(\left(x-5\right)^4+\left(x-2\right)^4=1^4+2^4=2^4+1^4\)

TH1 \(\left\{{}\begin{matrix}\left(x-5\right)^4=1^4\\\left(x-2\right)^4=2^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\\\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\)

TH2 \(\left\{{}\begin{matrix}\left(x-5\right)^4=2^4\\\left(x-2\right)^4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\end{matrix}\right.\)

Đặt x−72=ax−72=a. Khi đó PT trở thành:

(a−32)4+(a+32)4=17(a−32)4+(a+32)4=17

⇔2a4+27a2+818=17⇔2a4+27a2+818=17

⇔2a4+27a2=558⇔2a4+27a2=558

⇔a4+272a2=5516⇔a4+272a2=5516

⇔(a2+274)2=49⇔(a2+274)2=49

⇒[a2+274=7a2+274=−7<0(vô lý)⇒[a2+274=7a2+274=−7<0(vô lý)

⇒a2=14⇒a=±12⇒a2=14⇒a=±12

⇒x=a+72=[43

Đã cố gắng hết sức:(

3.3 a) x²

A)x²-(x-1)²=15

=>(x+x-1)(x-x+1)=15

=>(2x-1)1=15

=>2x-1=15

=>2x=16

=>x=8

B)16x-(4x-5)=15

=>(4x)-(4x-5)=15

=>(4x+4x-5)(4x-4x+5)=15

=>(8x-5)5=15

=>8x-5=3

=>8x=8

=>x=1

P = 1+ \(\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)

    = 1+ \(\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{x}{\left(x+2\right)\left(x-2\right)}-\dfrac{1}{x+2}\right)\)

    =1+\(\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x}{\left(x+2\right)\left(x-2\right)}-\dfrac{1.\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\)

   = 1 + \(\dfrac{1}{x+2}:\left(\dfrac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)

   =1 + \(\dfrac{1}{x+2}\): (\(\dfrac{6}{\left(x+2\right)\left(x-2\right)}\))

   =1 + \(\dfrac{1}{x+2}\).\(\dfrac{\left(x+2\right)\left(x-2\right)}{6}\)

   = 1 + \(\dfrac{x-2}{6}\)

  = \(\dfrac{6+x-2}{6}\)

= \(\dfrac{x+4}{6}\)

1. 5.(x+2)-2x.(x+2)=0

(x+2).(5-2x) = 0

+) TH1: x+2=0

=> x=-2

+) TH2: 5-2x=0

=>2x=5

=>x=5/2.

2. (x-1)^2 - 25=0

=> (x-1)^2 = 25

=> (x-1)^2 = 5^2

=> x-1 = 5 

=> x=6.

mn giúp mik vs , pls