Lời giải:

Lời giải:

Lời giải:

Lời giải:

Lời giải:

Lời giải:

Lời giải:

+ Vì 11 ≠ 0 (luôn đúng) nên phân thức $\frac{{\text{x}}^{\text{2}}-\text{9}}{\text{11}}$ luôn có nghĩa.

+ Ta có  

Vậy có hai giá trị của x thỏa mãn yêu cầu đề bài là:

x = 3; x = -3.

Lời giải:

Lời giải:

Lời giải:

Ta có:

$\frac{\text{5(y}-{\text{x)}}^{\text{2}}}{{\text{5x}}^{\text{2}}-\text{5xy}}=\frac{\text{5(x}-{\text{y)}}^{\text{2}}}{\text{5x(x}-\text{y)}}$

$=\frac{\text{x-y}}{\text{x}}$

$\Rightarrow \frac{\text{x}-\text{y}}{\text{x}}=\frac{\text{x}-\text{y}}{\text{P}}$

 => P = x.

Lời giải:

Lời giải:

Lời giải:

Ta có:

$\frac{{\text{5x}}^{\text{2}}-\text{13x}+\text{6}}{\text{A}}=\frac{\text{5x}-\text{3}}{\text{2x}+\text{5}}$

=> A.(5x - 3)

= (5x² - 13x + 6)(2x + 5)

A = (5x² - 13x + 6)(2x + 5) : (5x - 3)

= (5x² - 10x - 3x + 6)(2x + 5) : (5x - 3)

= [(5x(x - 2) - 3(x - 2)](2x + 5) : (5x - 3)

= (5x - 3)(x - 2)(2x + 5) : (5x - 3)

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

= 2x² + 5x - 4x - 10

= 2x² + x - 10.

Vậy A = 2x² + x - 10.

Lời giải:

Nhân cả tử và mẫu của phân thức đã cho với đa thức 3x(x+y)² ta được:

$\begin{array}{l}\frac{\text{x}+\text{y}}{\text{3x}}\\ =\frac{\text{(x}+\text{y).3x(x}+{\text{y)}}^{\text{2}}}{\text{3x.3x(x}+{\text{y)}}^{\text{2}}}\\ =\frac{\text{3x(x}+{\text{y)}}^{\text{3}}}{{\text{9x}}^{\text{2}}{\text{(x}+\text{y)}}^{\text{2}}}\end{array}$

Lời giải:

Ta có:

$\frac{\text{x}+\text{y}}{\text{3a}}=\frac{-\text{(x}+\text{y)}}{-\text{3a}}$

$=\frac{-\text{x}-\text{y}}{-\text{3a}}$ nên B, C sai.

Lại có

$\frac{\text{x}+\text{y}}{\text{3a}}=\frac{\text{(x}+\text{y).3a.(x}+\text{y)}}{\text{3a.3a.(x}+\text{y)}}$

$=\frac{\text{3a(x}+{\text{y)}}^{\text{2}}}{{\text{9a}}^{\text{2}}\text{(x}+\text{y)}}$ nên A sai, D đúng.

Lời giải:

Ta có:$\frac{{\text{x}}^{\text{2}}-\text{4x}+\text{3}}{{\text{x}}^{\text{2}}-\text{6x}+\text{9}}$

$=\frac{{\text{x}}^{\text{2}}-\text{3x}-\text{x}+\text{3}}{{\text{(x}-\text{3)}}^{\text{2}}}$

$=\frac{\text{x(x}-\text{3)}-\text{(x}-\text{3)}}{{\text{(x}-\text{3)}}^{\text{2}}}$

Lời giải:

Ta có

$\frac{\text{1}}{\text{x}-\text{y}}=\frac{{\text{1.(x}}^{\text{2}}-{\text{y}}^{\text{2}}\text{)}}{\text{(x}-{\text{y)(x}}^{\text{2}}-{\text{y}}^{\text{2}}\text{)}}$

$=\frac{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}{\text{(x}-\text{y)(x}-\text{y)(x}+\text{y)}}$

$=\frac{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}{{\text{(x}-\text{y)}}^{\text{2}}\text{(x}+\text{y)}}$

Lời giải:

Nhân cả tử và mẫu của phân thức  $\frac{\text{2}}{{\text{xy}}^{\text{3}}}$với x⁴y²(x - y) ta được:

Ta có:

$\frac{\text{2}}{{\text{xy}}^{\text{3}}}=\frac{{\text{2.x}}^{\text{4}}{\text{.y}}^{\text{2}}\text{(x}-\text{y)}}{{\text{xy}}^{\text{3}}{\text{.x}}^{\text{4}}{\text{y}}^{\text{2}}\text{(x}-\text{y)}}$

$=\frac{{\text{2x}}^{\text{5}}{\text{y}}^{\text{2}}-{\text{2x}}^{\text{4}}{\text{y}}^{\text{3}}}{{\text{x}}^{\text{5}}{\text{y}}^5\text{(x}-\text{y)}}$

Lời giải:

Nhân cả tử và mẫu của phân thức đã cho với số 3 ta được:

Ta có:

Lời giải:

Ta có

$-\frac{\text{x}-\text{3}}{\text{3}+\text{x}}=\frac{-\text{(x}-3\text{)}}{\text{3}+\text{x}}$

$=\frac{\text{3}-\text{x}}{\text{3}+\text{x}}$

*) $\frac{x^2-6x+9}{9-x^2}$

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

$=\frac{{(3-x)}^2:(3-x)}{(3-x)(3+x):(3-x)}$

*)$\frac{9-x^2}{{(3+x)}^2}$

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

$=\frac{(3-x)(3+x):(3+x)}{{(3+x)}^2:(3+x)}$

$=\frac{3-x}{3+x}$

*)$\frac{x-3}{3-x}=\frac{-(3-x)}{3-x}=-1$

Lời giải:

Ta có: $\frac{{\text{x}}^{\text{2}}-\text{3x}}{\text{9-3x}}$

$=\frac{-{\text{(x}}^{\text{2}}-\text{3x)}}{-\text{(9}-\text{3x)}}$

$=\frac{-{\text{x}}^{\text{2}}+\text{3x}}{\text{3x}-\text{9}}$ nên A đúng.

*$\frac{{\text{x}}^{\text{2}}-\text{3x}}{\text{9-3x}}=\frac{\text{x(x}-\text{3)}}{-\text{3(x}-\text{3)}}$

$=\frac{\text{x(x}-\text{3):(x}-\text{3)}}{-\text{3(x}-\text{3):(x}-\text{3)}}=\frac{-x}{\text{3}}$

≠ ${\frac{-x}{\text{3}}}^2$ nên B sai.

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

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

$=\frac{-x^2}{3}\ne \frac{-x}{3}$$=\frac{x^2-3x}{9-3x}$ nên C sai.

* $\frac{-{\text{2x}}^{\text{3}}}{\text{6x}}=\frac{-2x^3:2x}{6x:2x}$

$=\frac{-x^2}{3}\ne \frac{-\text{x}}{\text{3}}=\frac{{\text{x}}^{\text{2}}-\text{3x}}{\text{9-3x}}$ nên D sai.

Lời giải:

Ta có $\frac{\text{5x}+\text{5}}{\text{5x}}=\frac{\text{5(x}+\text{1)}}{\text{5x}}$

$=\frac{\text{5(x}+\text{1):5}}{\text{5x:5}}=\frac{\text{x}+\text{1}}{\text{x}}$

nên A đúng, D sai.

*) $\frac{{\text{x}}^{\text{2}}-\text{9}}{\text{x}+\text{3}}=\frac{\text{(x}-\text{3)(x}+\text{3)}}{\text{(x}+\text{3)}}$

$=\frac{\text{(x}-\text{3)(x}+\text{3):(x}+\text{3)}}{\text{(x}+\text{3):(x}+\text{3)}}=\text{x}-\text{3}$

nên B đúng.

*) $\frac{\text{x}+\text{3}}{{\text{x}}^{\text{2}}-\text{9}}=\frac{\text{x}+\text{3}}{\text{(x}+\text{3)(x}-\text{3)}}$

$=\frac{\text{(x}+\text{3):(x}+\text{3)}}{\text{(x}-\text{3)(x}+\text{3):(x}+\text{3)}}=\frac{\text{1}}{\text{x}-\text{3}}$

nên C đúng.

Lời giải:

+) Đáp án A: $\frac{\text{x-3}}{{\text{x}}^{\text{2}}-\text{9}}$

$=\frac{\text{x}-\text{3}}{\text{(x}-\text{3)(x}+\text{3)}}=\frac{\text{1}}{\text{x}+\text{3}}$

=> A đúng.

+) Đáp án B:$\frac{\text{3x-3}}{\text{3x}}$

$=\frac{\text{3(x}-\text{1)}}{\text{3x}}=\frac{\text{x}-\text{1}}{\text{x}}$

=> B đúng.

+) Đáp án C: $\frac{{\text{x}}^{\text{2}}-\text{6x}+\text{9}}{{\text{9-x}}^{\text{2}}}$

$=\frac{{\text{(3}-\text{x)}}^{\text{2}}}{\text{(3}-\text{x)(3}+\text{x)}}=\frac{\text{3}-\text{x}}{\text{x}+\text{3}}$

=> C đúng.

+) Đáp án D:$\frac{{\text{x(x}}^{\text{2}}-\text{4)}}{\text{2-x}}$

$=\frac{\text{x(x}-\text{2)(x}+\text{2)}}{\text{-(x-2)}}=\text{-x(x}+\text{2)}$

=> D sai.