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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) 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
\(A=\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}-\dfrac{x}{x-6}\)
\(=\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)
\(=\dfrac{12\left(x-3\right)}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}-\dfrac{x}{x-6}\)
\(=\dfrac{12}{2\left(x-6\right)}-\dfrac{x}{x-6}=\dfrac{6-x}{x-6}=-1\)
a: 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
ta có:
(\(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)
= (\(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)
= (\(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\)).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\). \(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{6}{x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{6}{x-6}\)- \(\dfrac{x}{x-6}\)
= \(\dfrac{6-x}{x-6}\)
= \(\dfrac{-\left(x-6\right)}{x-6}\)
= -1
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
Rút gọn biểu thức
a) \(A=\dfrac{x+3}{2x^2+6x}\)
\(A=\dfrac{1.\left(x+3\right)}{2x\left(x+3\right)}\)
\(A=\dfrac{1}{2x}\)
b) \(B=\dfrac{2x-9}{x-6}+\dfrac{2-x}{x-6}-\dfrac{1}{6-x}\)
\(B=\dfrac{2x-9}{x-6}+\dfrac{2-x}{x-6}+\dfrac{1}{x-6}\)
\(B=\dfrac{2x-9+2-x+1}{x-6}\)
\(B=\dfrac{x-6}{x-6}\)
\(B=1\)
a)S=\(\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)
=\(\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x\left(x+6\right)}+\dfrac{x}{6-x}\)
\(\left(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right):\dfrac{2x-6}{x\left(x+6\right)}+\dfrac{x}{6-x}\)
=\(\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)
=\(\dfrac{6\left(2x-6\right)x\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(2x-6\right)}+\dfrac{x}{6-x}\)
=\(\dfrac{6}{x-6}+\dfrac{x}{6-x}\)
=\(\dfrac{6}{x-6}-\dfrac{x}{x-6}=\dfrac{6-x}{x-6}=-1\)
b ) S khi rút gọn=-1 => mọi giá trị của x đều thỏa mãn S=-1