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\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)
\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
a) = x^2 - 9 - (x^2 + 3x - 10)
= -3x + 1
b) = 3x + 1 - 3x + 19
= 20
a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)
\(=x^2-9-x^2-3x+10\)
\(=-3x+1\)
b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)
\(=3x+1-3x+19\)
=20
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
a: \(=\dfrac{x^2-2x+1}{x}:\dfrac{x-1-3x^2+3x-3}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)
\(=\dfrac{\left(x-1\right)^3\cdot\left(x^2-x+1\right)}{-2x\left(x^2-2x+2\right)}\)
b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)
\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3+3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(1,\\ a,=2x^2+2x\\ b,=x^2+4x+3-4=x^2+4x-1\\ c,=x^2+4x+4+3x-5=x^2+7x-1\\ 2,\\ a,=3\left(x+y\right)\\ b,=\left(x-3\right)^2\\ c,=7\left(x+y\right)\\ 3,\\ \Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
bảo người ta ngu thì mày làm đi