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Ta có:
3 - x . 9 - x 2 = 3 - x 3 - x 3 + x = 3 - x 2 1)
Và 3 + x x 2 - 6 x + 9 = 3 + x . x - 3 2 = 3 + x . 3 - x 2 (2)
( vì ( x- 3) = - (3- x) nên x - 3 2 = - 3 - x 2 = 3 - x 2 )
Từ (1) và (2) suy ra: x - 3 . 9 - x 2 = 3 + x x 2 - 6 x + 9
Do đó: 
a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)
b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{-x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)
c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+2}{x-1}\)
d: \(\left(x^2-x-2\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2-3x+2\right)\left(x+1\right)\)
=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)
a) (x - 1)(x - 2). b) 4(x - 2)(x - 7).
c) (x + 2)(2x +1). d) (x - l)(2x - 7).
e) (2x + 3y - 3)(2x - 3y +1). g) (x - 3)( x 3 + x 2 - x +1).
h) (x + y)(x + y-l)(x + y + l).
Câu 1:
b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)
\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\dfrac{2}{x+3}\)
c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)
Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)
\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)
\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
\(\frac{x^2\left(x+2\right)}{x\left(x+2\right)^2}=\frac{x}{x+2}\Rightarrow\frac{x}{x+2}=\frac{x}{x+2}\)
\(\frac{3-x}{3+x}=\frac{x^2-6x+9}{9-x^2}\Rightarrow\frac{3-x}{3+x}=\frac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}\Rightarrow\frac{3-x}{3+x}=\frac{3-x}{3+x}\)
\(\frac{x^3-4x}{10-5x}=\frac{-x^2-2x}{5}\Rightarrow-\frac{x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\frac{-x^2-2x}{5}\)
\(\Rightarrow\frac{-x\left(x+2\right)}{5}=\frac{-x^2-2x}{5}\Rightarrow\frac{-x^2-2x}{5}=\frac{-x^2-2x}{5}\)
k nha bạn
sai rồi cái này là dùng định nghĩa 2 phân thức bằng nhau để chứng minh chúng bằng nhau mà
Ta có: x 3 - 4 x . 5 = 5 x 3 - 20 x
10 - 5 x - x 2 - 2 x = - 10 x 2 - 20 x + 5 x 3 + 10 x 2 = 5 x 3 - 20 x
Suy ra: x 3 - 4 x . 5 = 10 - 5 x - x 2 - 2 x
Vậy