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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: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
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
c/ đk: x khác 1; x khác -3
\(\dfrac{3x-1}{x-1}+\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)
\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)
\(\Leftrightarrow4x^2+11x+5=0\)
\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)
\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)
Vậy.........
d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đk: \(x\ne\pm\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)
\(\Leftrightarrow-102x-9=0\)
\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)
Vậy.........
a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)
\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)
\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)
\(\Leftrightarrow2x^3+4x^2+2x+24=0\)
\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)
\(\Leftrightarrow x\left(2x^2-2x+8\right)+3\left(2x^2-2x+8\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+8\right)\left(x+3\right)=0\)
\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)
Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)
=> x+ 3 = 0 <=> x= -3
Vậy......
b/ \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)
\(\Leftrightarrow x\left(2x^2+x+5\right)+\left(2x^2+x+5\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)
=> x + 1 = 0 <=> x = -1
Vậy....
\(a=\left(x-2\right)\left(x-4\right)\left(x^2-6x+10\right)\)
\(a=\left[x\left(x-4\right)-2\left(x-4\right)\right]\left(x^2-6x+10\right)\)
\(a=\left(x^2-4x-2x+8\right)\left(x^2-6x+10\right)\)
\(a=\left(x^2-6x+8\right)\left(x^2-6x+10\right)\)
\(a=\left(x^2-6x+9-1\right)\left(x^2-6x+9+1\right)\)
\(a=\left(x^2-6x+9\right)^2-1\ge-1\)
Dấu "=" xảy ra khi:
\(x^2-6x+9=0\)
\(\Rightarrow x^2-3x-3x+9=0\)
\(\Rightarrow x\left(x-3\right)-3\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)
\(b=\dfrac{20}{60x-9x^2-21}\)
\(b=\dfrac{20}{-9x^2+60x-21}\)
\(b=\dfrac{20}{-\left(9x^2-60x+21\right)}\)
\(b=\dfrac{20}{-\left(9x^2-60x+100-79\right)}\)
\(b=\dfrac{20}{-\left(9x^2-60x+100\right)+79}\)
\(b=\dfrac{20}{-\left(3x-10\right)^2+79}\le-\dfrac{20}{79}\)
Dấu "=" xảy ra khi: \(x=\dfrac{10}{3}\)