a) rút gọn

\(S=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

= \(\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

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

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

= \(\dfrac{6\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}\cdot\dfrac{x\left(x-6\right)\left(x+6\right)}{\left(2x-6\right)\left(x-6\right)}+\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}+\dfrac{-x}{-\left(6-x\right)}\)

= \(\dfrac{6}{x-6}+\dfrac{-x}{x-6}=\dfrac{6-x}{x-6}=-1\)

b)

Tìm x để giá trị của S = -1

Với mọi x khác 6 thì giá trị của S = -1

b)

Vì giá trị của biểu thức đã được xác định nên giá trị của

S = -1 không phụ thuộc vào giá trị của biến x.

a)

\(S=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

\(S=\left(\dfrac{x}{\left(x+6\right)\left(x-6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

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

\(S=\left(\dfrac{x^2-x^2+12x-36}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(S=\dfrac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(S=\dfrac{6}{x-6}-\dfrac{x}{x-6}\)

\(S=\dfrac{6-x}{x-6}=-1\)

b) Vì giá trị của biểu thức S không phụ thuộc vào giá trị của biến nên với mọi giá trị của x ta đều có giá trị của S là - 1.

Lời giải:

ĐKXĐ: \(x\neq -3; x\neq \pm 6; x\neq 0\)

Ta có:

\(A=\left(\frac{x}{x^2-36}-\frac{x+6}{x^2-6x}\right): \frac{2x+6}{x^2-6x}-\frac{x}{x+6}\)

\(A=\left(\frac{x}{x^2-36}-\frac{x+6}{x^2-6x}\right).\frac{x^2-6x}{2x+6}-\frac{x}{x+6}\)

\(=\frac{x(x^2-6x)}{(x^2-36)(2x+6)}-\frac{(x+6)(x^2-6x)}{x^2-6x)(2x+6)}-\frac{x}{x+6}\)

\(=\frac{x^2(x-6)}{(x-6)(x+6)(2x+6)}-\frac{x+6}{2x+6}-\frac{x}{x+6}\)

\(=\frac{x^2}{(x+6)(2x+6)}-\frac{(x+6)^2}{(2x+6)(x+6)}-\frac{x(2x+6)}{(2x+6)(x+6)}\)

\(=\frac{x^2-(x+6)^2-x(2x+6)}{(x+6)(2x+6)}=\frac{-(2x^2+18x+36)}{2x^2+18x+36}=-1\)

a: ĐKXĐ: x<>0; x<>6; x<>-6;x<>3

\(A=\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2\left(x-3\right)}{x\left(x+6\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{x^2-x^2+12x-36}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{12\left(x-3\right)}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

b: Khi \(x\in R\backslash\left\{0;6;-6;3\right\}\) thì A luôn bằng -1

c: Khi x=1 thì A=-1

\(A=\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}-\dfrac{x}{x-6}\)

\(=\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{12\left(x-3\right)}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{12}{2\left(x-6\right)}-\dfrac{x}{x-6}=\dfrac{6-x}{x-6}=-1\)

a) Tớ làm luôn nhé , không chép lại đề đâu

P = \(\left[\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right].\dfrac{x\left(x+6\right)}{2x-6}\)

ĐKXĐ : x # -6 ; x # 6 ; x # 0 ; x # 3 . Khi đó , ta có :

P = \(\left[\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]\).\(\dfrac{x\left(x+6\right)}{2x-6}\)

P = \(\dfrac{x^2-x^2+12x-36}{x-6}.\dfrac{1}{2x-6}\)

P = \(\dfrac{6\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}=\dfrac{6}{x-6}\)

b) Tương tự

a) ĐKXĐ: \(x\ne0;x\ne6;x\ne-6\)

b) \(A=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

\(A=\left[\dfrac{x}{\left(x+6\right)\left(x-6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right]:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\left[\dfrac{x^2}{x\left(x+6\right)\left(x-6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right]:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{x^2-x^2+12x-36}{x\left(x+6\right)\left(x-6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{12x-36}{x\left(x+6\right)\left(x-6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}-\dfrac{x}{x-6}\)

\(A=\dfrac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(A=\dfrac{6}{x-6}-\dfrac{x}{x-6}\)

\(A=\dfrac{6-x}{x-6}\)

\(A=-\dfrac{x-6}{x-6}\)

\(A=-1\)

Vậy giá trị của A không phụ thuộc vào giá trị của biến 

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

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

b: ĐKXĐ: x<>1

\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)

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

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

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

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

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

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

e: ĐKXĐ: x<>-y

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

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

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

\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

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

\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

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

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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