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\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)
b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)
a: ĐKXĐ: x<>0; x<>-1
PT =>x+1-2x=3
=>1-x=3
=>x=-2(nhận)
b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)
=>x-3=5(2x-3)
=>10x-15=x-3
=>9x=12
=>x=4/3(nhận)
c: ĐKXĐ: x<>0; x<>2
PT =>x(x+2)-x+2=2
=>x^2+2x-x=0
=>x(x+1)=0
=>x=-1
a: =>4x^2-2x-x-4=0
=>4x^2-3x-4=0
=>\(x=\dfrac{3\pm\sqrt{73}}{8}\)
b: =>x^2+x-x+2=5x-4
=>x^2+2=5x-4
=>x^2-5x+6=0
=>x=2(loại) hoặc x=3(nhận)
1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
2: \(C=A:B\)
\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)
\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)
=>C>=-1
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3-x^2-2x^2+4x+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3+x^2+4x}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+x+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
đk : x khác 0 ; 1 ; 2
\(2\left(x-1\right)\left(x-2\right)-x\left(x-2\right)=2x\left(x-1\right)\)
\(\Leftrightarrow2\left(x^2-3x+2\right)-x^2+2x=2x^2-2x\)
\(\Leftrightarrow-6x+4-x^2+2x=-2x\Leftrightarrow x^2+2x-4=0\)
\(\Leftrightarrow\left(x+1\right)^2-5=0\Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{5}\\x=-1+\sqrt{5}\end{matrix}\right.\left(tm\right)\)
Thấy hơi lạ lạ, mấy bạn coi rồi nói cho mình nha! Cảm ơn~