1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

a) \(x=\dfrac{25}{72}\)

b)\(x=-\dfrac{1}{4}\)

  \(x=\dfrac{3}{2}\)

c)\(x=\dfrac{5}{4}\) hoặc

  x \(=\dfrac{8}{5}\)

d và e chịu vì mk kg giỏi lắm về mũ 

f)\(x=-2\)

G)\(x=-\dfrac{5}{12}\)

a: =91/105+60/105-101/105

=50/105=10/21

c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)

d: =2-2/9

=18/9-2/9

=16/9

e: =24/36-9/36+8/36

=23/36

g: =5/2+1/2

=3

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 $x=1$

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.

mọi người ơi giúp với tui đang cần gấp .Ai làm nhanh nhất thì tui sẽ h cho

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: \(S=\varnothing\)

\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow3x+6+2x+2=5x+4\)

\(\Leftrightarrow3x+2x-5x=-6-2+4\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm 

\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow4x-2-15=9x-3\)

\(\Leftrightarrow4x-9x=2+15-3\)

\(\Leftrightarrow-5x=14\)

.....

mấy cái này mẫu nào dài cậu phân tích ra : 

VD : câu  3 : \(3x^2-4x+1\)

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

\(=3x\left(x-1\right)-\left(x-1\right)\)

\(=\left(3x-1\right)\left(x-1\right)\)

r bắt đầu giải PHương trình :)) Mấy câu còn lại tương tự 

4; \(\frac{5}{x-2}+\frac{2}{x+4}=\frac{3x}{x^2+2x-8}.\)

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

\(\Leftrightarrow5x+20+2x-4=3x\)

\(\Leftrightarrow4x=-16\Leftrightarrow x=-2\left(TM\right)\)

KL ::

\(5;\frac{4}{x+6}+\frac{1}{x-3}=\frac{9}{x^2+3x-18}\)

\(\Leftrightarrow\frac{4\left(x-3\right)}{\left(x+6\right)\left(x-3\right)}+\frac{x+6}{\left(x-3\right)\left(x+6\right)}=\frac{9}{\left(x-3\right)\left(x+6\right)}\)

\(\Leftrightarrow4x+x=3+9-6\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\)