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f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1
=>-16x-34=x-1
=>-17x=33
=>x=-33/17
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6
=>4x^2+16x-20-4x^2-10x+4=0
=>6x=16
=>x=8/3
a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y
=5x^3-7x^2y+2xy^2+5x-2y
b: =(x^2-1)(x+2)
=x^3+2x^2-x-2
c: =1/2x^2y^2(4x^2-y^2)
=2x^4y^2-1/2x^2y^4
d: =(x^2-1/4)(4x-1)
=4x^3-x^2-x+1/4
e: =x^2-2x-35+(2x+1)(x-3)
=x^2-2x-35+2x^2-6x+x-3
=3x^2-7x-38
1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
\(\dfrac{1}{x}+\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}\)
=2/x
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{-1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow-\left(x-3\right)=2010\)
\(\Rightarrow x=-2013\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow x=2007\)
x=-1
x=2
\(x+\frac{1}{2}=\frac{1}{x}+1\) (điều kiện \(x\ne0\))
\(\Leftrightarrow x-\frac{1}{x}=1-\frac{1}{2}\)\(\Leftrightarrow\frac{x^2-1}{x}=\frac{1}{2}\)\(\Rightarrow x^2-1=\frac{x}{2}\)\(\Leftrightarrow x^2-\frac{x}{2}-1=0\)
\(\Leftrightarrow x^2-\frac{x}{2}+\frac{1}{16}-\frac{17}{16}=0\)\(\Leftrightarrow x^2-2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2-\left(\sqrt{\frac{17}{16}}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{4}\right)^2-\left(\frac{\sqrt{17}}{4}\right)^2=0\)\(\Leftrightarrow\left(x-\frac{1}{4}-\frac{\sqrt{17}}{4}\right)\left(x-\frac{1}{4}+\frac{\sqrt{17}}{4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{4}-\frac{\sqrt{17}}{4}=0\\x-\frac{1}{4}+\frac{\sqrt{17}}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{17}}{4}\\x=\frac{1-\sqrt{17}}{4}\end{cases}}\)(nhận)
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