X * 2012 - X=2012*2010+2012
x-2012/2008-x-2012/2009=x-2012/2010-x-2012/2011.tìm x
y x 2012 - y = 2012 x 2010 + 2012
y x 2012 - y = 2012 x 2010 + 2012
2011 x y = 4046132
y = 4046132 : 2011
y = 2012
vậy y = 2012
cbht
Ta có:
y x 2012 - y = 2012 x 2010 + 2012
y*2012 - y*1=2012*2010+2012*1
y*(2012- 1)=2012*(2010+1)
y*2011=2012*2011
y=2012(bớt mỗi vế 1 thừa số 2011)
Vậy y=2012
TL:
y x 2012 - y = 2012 x 2010 + 2012
<=> 2011y = 2012 x 2011
<=> y = 2012 x 2011 : 2011
<=> y = 2012
#study well#
Tìm X biết X x 2012 - X = 2012 x 2010 + 2012
X x 2012 - X = 2012 x 2010 + 2012
X x ( 2012 - 1 ) = 2012 x ( 2010 + 1 )
X x 2011 = 2012 x 2011
\(\Rightarrow\)X = 2012
X x 2012 - X = 2012 x 2010 + 2012
<=>2011xX=2012x2011
<=>X=2012
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
Cho x=2011.Tính GTBT:
A= \(x^{2011}-2012.x^{2010}+2012.x^{2009}-2012.x^{2008}+...-2012.x^2+2012.x-1^{ }\)
Ta có: x=2011 \(\Rightarrow\)x+1=2012
\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)
=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)
= \(x-1=2011-1=2010\)
=
Thay 2012=x+1.
\(A=x^{2011}-\left(x+1\right)x^{2010}+\left(x+1\right)x^{2009}-\left(x+1\right)x^{2008}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(A=x^{2011}-x^{2011}-x^{2010}+x^{2010}+x^{2009}-...-x^3-x^2+x^2+x-1\)
\(A=x-1=2011-1=2010\)
x^2010+y^2010=x^2011+x^2011=x^2012+y^2012. tính x^2016+y^2016
|x-2010|+|x-2012|=2012
Giải pt sau : \(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
Lời giải:
Ta có:
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow \left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow \frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow (x-2013)\left(\frac{1}{2012}+\frac{1}{2011}+...+1\right)=0\)
Dễ thấy \(\frac{1}{2012}+\frac{1}{2011}+...+1\neq 0\Rightarrow x-2013=0\)
\(\Leftrightarrow x=2013\)
Vậy PT có nghiệm \(x=2013\)
giải phương trình: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x+3}{2010}+...+\frac{x-2012}{1}=2012\)
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=2012\)
\(\Leftrightarrow\)\(\frac{x-1-2012}{2012}+\frac{x-2-2011}{2011}+\frac{x-3-2010}{2010}+...+\frac{x-2012-1}{1}=0\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\ne0\)
Nên \(x-2013=0\)
\(\Leftrightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt ~
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1+2012=2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
\(\Leftrightarrow x=2013\)
x x 2012 - x = 2012 x2010 + 2012
giúp tớ tớ với nha
\(x.2012-x=2012.2010+2012\)
\(2012x-x=2012.2010+2012\)
\(2011x=4046132\)
\(x=4046132:2011=2012\)
Vậy x=2012
Lần sau bn nên ghi đề rõ hơn , dấu nhân bn cứ thay = dấu "." là đc r
\(x.2012-x=2012.2010+2012\)
\(x.2012-x.1=2012.2010+2012.1\)
\(x.\left(2012-1\right)=2012.\left(2010+1\right)\)
\(x.2011=2012.2011\)
\(\Rightarrow x=2012\)
~ Hok tốt ~