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con A bn bấm nhầm đúng ko mik sửa lại nhé
A= 20142 - 4018. 1014 + 10142
= (2014 - 1014)2
= 10002
= 1000000
B= 9x2 - 6x + 2013
= 9x2 - 6x + 1 + 2012
= (3x - 1)2 + 2012
thay x = \(\dfrac{200001}{3}\)vào biểu thức B ta có:
B = (3.\(\dfrac{200001}{3}\)- 1)2 + 2012
= (200001 - 1)2 + 2012
= 2000002 + 2012
= 40000002012
mik chỉ làm đc đến đây thôi nhưng mong bn ủng hộ! 
\(x\left(x+5\right)=9x\)
\(\Leftrightarrow xx+5x=9x\)
\(\Leftrightarrow xx+5x-9x=0\)
\(\Leftrightarrow xx-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=4\)
1: \(MTC=2\left(x-y\right)\left(x+y\right)\)
\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{x+y}{2x^2+4xy+2y^2}\)
\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)
\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)
2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
\(x^2+3y^2=4xy\)
\(\Leftrightarrow x^2-4xy+3y^2=0\)
=>(x-y)(x-3y)=0
=>x=y hoặc x=3y
Khi x=y thì \(A=\dfrac{2\cdot y+2013y}{y-2y}=-2015\)
Khi x=3y thì \(A=\dfrac{2\cdot3y+2013y}{3y-2y}=2019\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
a \(=9x^2-6x+1+2012\)
\(=\left(3x-1\right)^2+2012\)
\(=200000^2+2012\)
b: \(=2014^2-2\cdot2014\cdot1014+1014^2\)
\(=\left(2014-1014\right)^2=1000^2=10^6\)
c: \(x^2+3y^2=4xy\)
=>x^2-4xy+3y^2=0
=>(x-y)*(x-3y)=0
=>x=y hoặc x=3y
KHi x=y thì \(C=\dfrac{2x+2013x}{x-2x}=-2015\)
Khi x=3y thì \(C=\dfrac{6y+2013y}{3y-2y}=2019\)