2: \(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)

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

=>(x+3)(x-2)=0

=>x=-3 hoặc x=2

5: \(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)

hay \(x\in\left\{-2;1;-1\right\}\)

7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -2

=> -11x + (-39) = -2

=> -11x = -2 - (-39)

=> -11x = 37

=> x = \(\frac{-37}{11}\)

7(2x - 5) - 5(7x - 2) + 2(5x - 7)             = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14         = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -6
=> -11x - 39                                         = -6
=> -11x                                                = -6+39
=> -11x                                                = 33
=>      x                                                = 33:(-11)
=>      x                                                = -3

chỉ đoán mò được x=-1, chứ không biết giải huhu, còn nghiệm nào nữa không, các bạn giải giúp mình với...

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

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

\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)

\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)

\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)

đùa ak

1.a)x+7=-5-14

x+7=-19

x=-19-7

x=-26

Vậy x=-26

b)3x-4=(-2)³-11

3x-8=-11

3x=-11+8

3x=-3

x=-3:3

X=-1

Vậy x=-1

c)11+(4x-11)=-9-(-15)

11+(4x-11)=-9+15

11+(4x-11)=6

4x-11=6-11

4x=6

x=\(\dfrac{6}{4}\)

Vậy x=\(\dfrac{6}{4}\)

d)\(\left|2x-1\right|=5\)

⇒2x-1=\(\pm5\)

+Nếu 2x-1=5

2x=6

x=6:2

x=3

Vậy x=3

+Nếu 2x-1=-5

2x=-5+1

2x=-4

x=-4:2

x=-2

Vậy xϵ{3;-2}

e)\(\left|x-7\right|+3=25\)

\(\left|x-7\right|=22\)

⇒x-7=\(\pm22\)

+Nếu x-7=22

x=22+7=29

+Nếu x-7=-22

x=-22+7

x=-15

Vậy xϵ{29;-15}

\(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)

\(\Leftrightarrow x^3+64-x\left(x^2-25\right)=264\)

\(\Leftrightarrow x^3+64-x^3+25x=264\)

\(\Leftrightarrow25x=200\)

\(\Leftrightarrow x=8\)