a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

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

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

Suy ra: \(x^2-1+x-2x+1=0\)

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

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow12x=-6\)

hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.

Sửa lại môn học để các bạn làm nhé em!

bạn sửa lại môn hôn học đi ạ

14) Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{2x+4}{x^2-2x-3}\)

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

Suy ra: \(x^2+x+x^2-3x-4x-8=0\)

\(\Leftrightarrow2x^2-6x-8=0\)

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

a=1; b=-3; c=-4

Vì a-b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=-1\left(loại\right);x_2=\dfrac{-c}{a}=4\left(nhận\right)\)

9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)

\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)

\(\Leftrightarrow-4x=9\)

hay \(x=-\dfrac{9}{4}\)

10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}

11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)

Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)

\(\Leftrightarrow5x^2-7x=0\)

\(\Leftrightarrow x\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

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

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

Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)

\(\Leftrightarrow2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)

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

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

Suy ra: \(x^2+2x-x+2-2=0\)

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

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

p) \(\left(9-x\right)\left(x^2+2x-3\right)\)

\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)

\(=9x^2+18x-27-x^3-2x^2+3x\)

\(=-x^3+7x^2+21x-27\)

n) \(\left(-x+3\right)\left(x^2+x+1\right)\)

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

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

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

o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)

\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)

\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)

\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)

q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)

\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)

\(=6x^3-12x^2-18x+x^2-2x-3\)

\(=6x^3-11x^2-20x-3\)

r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)

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

\(=-2x^3-6x^2+2x-x^2-3x+1\)

\(=-2x^3-7x^2-x+1\)

u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)

\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)

\(=-2x^3+2x^2+12x+3x^2-3x-18\)

\(=-2x^3+5x^2+9x-18\)

s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)

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

\(=-4x^3-12x^2+8x+5x^2+15x-10\)

\(=-4x^3-7x^2+23x-10\)

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)

\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)

\(=-x^2-3+2x^4+6x+18-12x^3\)

\(=2x^4-12x^3-x^2+6x+15\)

p: (-x+9)(x^2+2x-3)

=-x^3-2x^2+3x+9x^2+18x-27

=-x^3+7x^2+21x-27

n: (-x+3)(x^2+x+1)

=-x^3-x^2-x+3x^2+3x+3

=-x^3+2x^2+2x+3

o: (-6x+1/2)(x^2-4x+2)

=-6x^3+24x^2-12x+1/2x^2-2x+1

=-64x^3+49/2x^2-14x+1

q: (6x+1)(x^2-2x-3)

=6x^3-12x^2-18x+x^2-2x-3

=6x^3-11x^2-20x-3

r: (2x+1)(-x^2-3x+1)

=-2x^3-6x^2+2x-x^2-3x+1

=-2x^3-7x^2-x+1

u: =-2x^3+2x^2+12x+3x^2-3x-18

=-2x^3+5x^2+9x-18

s: =-4x^3-12x^2+8x+5x^2+15x-10

=-4x^3-7x^2+23x-10

c: \(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\)

\(=4x^2+12x+9+4x^2-12x+9-\left(4x^2-9\right)\)

\(=8x^2+18-4x^2+9=4x^2+27\)

d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)

\(=x^3-1-8x^3-27=-7x^3-28\)

e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)

\(=x^3+3x^2+3x+1-6x^2-\left(x^3-3x^2+3x-1\right)\)

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

=2

1.A =( x-3)( x+3) + 15 - x²

   A=X²-3X+3X+15-X³

  A=15-X

2.B=(X -1) (X²+X+1) - X (X²+2) + 2X  

 B=X³+ X²+ X - X² - X - 1 - X³ - 2X + 2X

B=   -1

3.C=(2X - 1 ) (4X² + 2X + 1) - X ( 8 X 2 + 1 ) + X

C=8X³ - 4X² +4X² - 2X +2 X - 1 - 8X²2 - X + X

C=8X³ - 1 - 8X²2

MK CHỈ LM ĐC TỚI ĐÓ THUI SAI CHỖ NÀO ĐỪNG TRÁCH VÌ MK YẾU PHẦN NÀY

Lời giải:
a.

a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$

$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$

$\Leftrightarrow 3x+1=5x-3$

$\Leftrightarrow 4=2x$

$\Leftrightarrow x=2$

b.

$(2x-1)(x+3)-(x-2)(x+3)=3x+1$

$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$

$\Leftrightarrow x^2+5x+1=3x+1$

$\Leftrightarrow x^2+2x=0$

$\Leftrightarrow x(x+2)=0$

$\Leftrightarrow x=0$ hoặc $x=-2$

c.

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

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

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

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

$\Leftrightarrow x-1=0$ hoặc $-x=0$

$\Leftrightarrow x=1$ hoặc $x=0$

d.

$4x(x-5)-(2x-3)(2x+3)=9$

$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$

$\Leftrightarrow -20x=0$

$\Leftrightarrow x=0$

a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)

\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow2x=4\)

hay x=2

b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)

\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)

\(\Leftrightarrow4x^2-20x-4x^2+9=9\)

hay x=0

\(\dfrac{2x-1}{x+1}=\dfrac{-2x+1}{x-5}\left(x\ne-1;5\right)\)

\(\dfrac{2x-1}{x+1}=\dfrac{2x-1}{5-x}\)

\(x+1=5-x\)

\(2x=4\Rightarrow x=2\)