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\(3\left(x+1\right)^2+2\left(x-3\right)^3-5\left(x-5\right)\left(x+5\right)\)
\(=3\left(x^2+2x+1\right)+2\left(x^2-6x+9\right)-5\left(x^2-25\right)\)
\(=3x^2+6x+3+2x^2-12x+18-5x^2+125\)
\(=-6x+146\)
\(\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}\)
\(=\dfrac{x^3\left(x^2-3x+5\right)-\left(x^2-3x+5\right)}{x^2-3x+5}\)
\(=x^3-1\)
a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)
\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)
\(=7x-16\)
2:
a: =>-2x=10
=>x=-5
b: =>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{3x+4}\cdot\dfrac{-5}{x-5}=\dfrac{-5\left(x+5\right)}{3x+4}\)
`a, 20x^3y^5 : 5x^2y^2`
`= (20:5)x^(3-2) . y^(5-2)`
`= 4xy^3`
`b, 18x^3y^5 : (3(-x^3)y^2)`
`= -(18:3)y^(5-3)`
`= -6y^2`
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
Bài 3:
\(\Leftrightarrow x^3+64-x^3+25x=264\)
hay x=8
\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)
