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a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)
\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)
=2x^2-3x+4
b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)
\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)
c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
\(\left(x^2+4x+8\right)\left(x^2+5x+8\right)=2x^2\left(1\right)\)
\(\Leftrightarrow x^4+5x^3+8x^2+4x^3+20x^2+32x+8x^2+40x+64-2x^2=0\)
\(\Leftrightarrow x^4+5x^3+4x^3+8x^2+20x^2+8x^2-2x^2+40x+32x+64=0\)
\(\Leftrightarrow x^4+9x^3+34x^2+72x+64=0\)
\(\Leftrightarrow x^4+2x^3+7x^3+14x^2+20x^2+40x+32x+64=0\)
\(\Leftrightarrow x^3\left(x+2\right)+7x^2\left(x+2\right)+20x\left(x+2\right)+32\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+7x^2+20x+32\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+4x^2+3x^2+12x+8x+32\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x+4\right)+3x\left(x+4\right)+8\left(x+4\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+3x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\\x^2+3x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\\vô.nghiệm\left(\Delta=9-32=-23< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\) là nghiệm của phương trình \(\left(1\right)\)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
Ta có \(2y^2⋮2\Rightarrow x^2\equiv1\left(mod2\right)\Rightarrow x^2\equiv1\left(mod4\right)\Rightarrow2y^2⋮4\Rightarrow y⋮2\Rightarrow x^2\equiv5\left(mod8\right)\) (vô lí).
Vậy pt vô nghiệm nguyên.
2: \(PT\Leftrightarrow3x^3+6x^2-12x+8=0\Leftrightarrow4x^3=\left(x-2\right)^3\Leftrightarrow\sqrt[3]{4}x=x-2\Leftrightarrow x=\dfrac{-2}{\sqrt[3]{4}-1}\).