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a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

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

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

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

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

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

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

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

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)

\(=\dfrac{-3}{x-3}\)

giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn

a) x²+\(\dfrac{2x}{x-1}\)=8(ĐKXĐ : x ≠ 1

⇔ x²(x-1)+2x =8⇔ x³ - x² +2x - 8=0

⇔x³ - 2³ -x²+2x =0⇔ (x-2)(x² +x+1) -x(x-2)

⇔(x-2)(x² +1)⇒x =2

x² +1 =0⇒x² -1⇒x ∈∅(loại)

vậy x =2

Giải phương trình :

a) \(x^2+\dfrac{2x}{x-1}=8\)

ĐKXĐ : \(x-1\ne0\Rightarrow x\ne1\)

Ta có : \(x^2+\dfrac{2x}{x-1}=8\)

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

\(\Rightarrow x^2\left(x-1\right)+2x=8\left(x-1\right)\)

\(\Leftrightarrow x^3-x^2+2x=8x-8\)

\(\Leftrightarrow x^3-x^2+2x-8x=-8\)

\(\Leftrightarrow x^3-x^2-6x+8=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x-8\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-2\left(3x-4\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)

Đối chiếu với ĐKXĐ ta được \(x\in\left\{\sqrt{2};\dfrac{4}{3}\right\}\) thỏa mãn.

a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)

b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

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

c: \(=6x-y+2x^2+3y-2x^2+x\)

\(=7x+2y\)

d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

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

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị