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a: \(\dfrac{x-1}{x^2-x+1}-\dfrac{x+1}{x^2+x+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)=10\)

\(\Leftrightarrow x\left(x^3-1\right)-x\left(x^3+1\right)=10\)

=>-2x=10

hay x=-5

d: \(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+7\right)\left(x+8\right)}=\dfrac{1}{14}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}\)

\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=14\left(x+8\right)-14\left(x+1\right)\)

\(\Leftrightarrow x^2+9x+8=14x+112-14x-14=98\)

\(\Leftrightarrow x^2+9x-90=0\)

\(\Leftrightarrow x\in\left\{6;-15\right\}\)

Let's solve each equation step by step:

Squaring both sides of the equation, we get:
x^2 - 6x + 9 = (3 - x)^2
x^2 - 6x + 9 = 9 - 6x + x^2

The x^2 terms cancel out, and we are left with:
-6x = -6x

This equation is true for any value of x. Therefore, there are infinitely many solutions.

Moving all terms to one side of the equation, we get:
x^2 - (1/2)x - x + 3/2 - 1/16 = 0
x^2 - (3/2)x + 29/16 = 0

To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -3/2, and c = 29/16. Plugging in these values, we get:
x = (3/2 ± √((-3/2)^2 - 4(1)(29/16))) / (2(1))
x = (3/2 ± √(9/4 - 29/4)) / 2
x = (3/2 ± √(-20/4)) / 2
x = (3/2 ± √(-5)) / 2

Since the square root of a negative number is not a real number, this equation has no real solutions.

Squaring both sides of the equation, we get:
(x - 2)(x - 1) = (x - 1) - 2√(x - 1) + 1
x^2 - 3x + 2 = x - 1 - 2√(x - 1) + 1
x^2 - 4x + 2 = -2√(x - 1)

Squaring both sides again, we get:
(x^2 - 4x + 2)^2 = (-2√(x - 1))^2
x^4 - 8x^3 + 20x^2 - 16x + 4 = 4(x - 1)
x^4 - 8x^3 + 20x^2 - 16x + 4 = 4x - 4

Rearranging terms, we have:
x^4 - 8x^3 + 20x^2 - 20x + 8 = 0

This equation does not have a simple solution and requires further calculations or approximation methods to find the solutions.

Simplifying the left side of the equation, we get:
3 - 4√5 - √5 = -2
-√5 - 5 = -2
-√5 = 3

This equation is not true since the square root of a number cannot be negative.

Therefore, the given equations either have infinitely many solutions or no real solutions.

Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

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

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

$x(x-3)-x+3=0$

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

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

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

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

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

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

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)