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\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
\(a.\)
\(P=\left[\left(\dfrac{1}{x^2}+1\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\left(\dfrac{1}{x^2}+\dfrac{x^2}{x^2}\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+\dfrac{x}{x}\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^2}.\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{x+1}{x}\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^2\left(x^2+2x+1\right)}+\dfrac{2}{x\left(x+1\right)^2}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2}{x^3+2x^2+x}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x\left(x^3+2x^2+x\right)}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x^4+2x^3+x^2}\right].\dfrac{x-1}{x^3}\)
\(P=\dfrac{x^2+1+2x}{x^4+2x^3+x^2}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{x^2+2x+1}{x^2\left(x^2+2x+1\right)}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{1}{x^2}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{x-1}{x^5}\)
Chia đa thức cho đơn thức
\(4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\)
=\(\left(3x-4\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\)
= \(\left(3x-4+12x^2-3x\right):\left(-3x-2x-1\right)\)
\(=\left(-4+12x^2\right):\left(-5x-1\right)\)
=\(\dfrac{-4+12x^2}{1}.\dfrac{-1}{5x-1}\)
=\(\dfrac{-4+12x^2}{5x}=\dfrac{-4+12x}{5}\)
a)
(x-2y)2 >= 0 V x,y
(y-2018)>=0 V y
=> P=(ghi lại đề) >= 0
vậy GTNN của p bằng 0
dấu "=" xảy ra (=) \(\hept{\begin{cases}x-2y=0\\y-2018=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\y=2018\end{cases}}\left(=\right)\hept{\begin{cases}y=2018\\x=4036\end{cases}}\)
b) (x+y-3)4 >= 0 V x,y
(x-2y)2 >= V x,y
=> Q=(ghi lại đề) >= 2018
vậy GTNN của Q bằng 2018
dấu "=" xảy ra (=) \(\hept{\begin{cases}x+y-3=0\\x-2y=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\3y=3\end{cases}}\left(=\right)\hept{\begin{cases}y=1\\x=2\end{cases}}\)
c)
(2x + 1/6)4>= 0 V x
=> N=(ghi lại đề) >= -2
vậy GTNN của N bằng -2
dấu "=" xảy ra (=) 2x+1/6=0
(=) 2x=-16
(=) x=-1/12
#Học-tốt
\(=\left(\dfrac{x+1-1}{x+1}\right)\left(\dfrac{x+2-1}{x+2}\right)...\left(\dfrac{x+2014-1}{x+2014}\right)\)
\(=\dfrac{x\left(x+1\right)\left(x+2\right)...\left(x+2013\right)}{\left(x+1\right)\left(x+2\right)...\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{x}{x+2014}\)
\(C=\left(x^3+y^3\right)+3xy\left(x^2+y^2+2xy\left(x+y\right)\right)\)
\(C=\left(x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2\right)+3xy\left(x^2+y^2+2xy\right)\) (vì x+y=1)
\(C=\left(x+y\right)^3-3x^2y-3xy^2+3xy\left(x+y\right)^2\)
\(C=1^3-3xy\left(x+y\right)+3xy.1^2\) (vì x+y=1)
\(C=1-3xy+3xy\)(vì x+y=1)
\(C=1\)
\(D=2\left(\left(x+y\right)^3-3xy\left(x+y\right)\right)-3\left(\left(x+y\right)^2-2xy\right)\)
\(D=2\left(1^3-3xy\right)-3\left(1^2-2xy\right)\)(vì x+y=1)
\(D=2-6xy-3+6xy\)
\(D=-1\)
\(=\dfrac{x^2-2x+1}{4}+x^2-1+x^2+2x+1\)
\(=\dfrac{x^2-2x+1+4\left(2x^2+2x\right)}{4}\)
\(=\dfrac{9x^2+6x+1}{4}\)
\(=\dfrac{\left(3x+1\right)^2}{4}\)