\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)

\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)

\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)

\(B=0\)

Phúc 6A phải k

xét các thừa số tích B có: \(1-\frac{2010}{2010}=0\)

Nên B = 0

a)(3/2-0,5)/x=7/2+1/4

(3/2-1/2)/x=14/4+1/4

1/x=15/4

x=1:15/4

x=4/15

b)(x*0,25+2010)*2011=(53+2010)*(2012-1)

(x*0,25+2010)*2011=2063*2011

=>0,25x+2010=2063

0,25x=2063-2010

0,25x=53

x=53/0,25

x=212

`(2011xx2020-1)/(2009xx2011+2010)`

`=((2009+1)xx2011-1)/(2009xx2011+2010)`

`=(2009xx2011+2011-1)/(2009xx2011+2010)`

`=(2009xx2011+2010)/(2009xx2011+2010)`

`=1`

\(\dfrac{2011.2010-1}{2009.2011+2010}\) 

= \(\dfrac{2011.2009+2011-1}{2009.2011+2010}\) 

= \(\dfrac{2011.2009+2010}{2009.2011+2010}\) 

= 1

A = 2011 x (2009  + 1) - 1/ 2011 x 2009 + 2010

A = 2011 x 2009 + 2011 x 1 - 1/2011 x 2009 + 2010

A = 2011 x 2009 + 2010/2011 x 2009 + 2010

A = 1

B.

Từ \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)

\(\Rightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)

\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\left(1\right)\)

*)Xét \(x+y+z\ne0\left(2\right)\). Từ (1) và (2)

\(\Rightarrow x=y=z\). Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=2\cdot2\cdot2=8\)

*)Xét \(x+y+z=0\)\(\Rightarrow\left\{\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=\frac{-z}{y}\cdot\frac{-x}{z}\cdot\frac{-y}{x}=-1\)

a)

Ta có \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)

\(\Rightarrow\left\{\begin{matrix}\frac{y+z-x}{x}=1\\\frac{z+x-y}{y}=1\\\frac{x+y-z}{z}=1\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\) (1)

Ta có \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)

\(\Rightarrow B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)

Thế (1) vào biểu thức B

\(\Rightarrow B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}\)

\(\Rightarrow B=2.2.2=8\)

Vậy biểu thức \(B=8\)

b) Theo mình bằng 4

Chọn A

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(=>\dfrac{x+4}{2010}+1\))+(\(\dfrac{x+3}{2011}+1\))=\(\left(\dfrac{x+2}{2012}+1\right)\)+\(\left(\dfrac{x+1}{2013}+1\right)\)

=>\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

=>x+2014(\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\))=0

ta thấy \(\dfrac{1}{2010}>\dfrac{1}{2011}>\dfrac{1}{2012}>\dfrac{1}{2013}\)

=>\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}>0\)

để A=0

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow\)x=-2014

a)\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)Mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

tìm x biết :

a)(x+4) / 2010 + (x+3)/2011 =( x+2)/2012 + (x+1)/2013

==> [(x+4)/2010 + 1] + [(x+3)/2011 + 1] =[( x+2)/2012 + 1] + [(x+1)/2013 + 1].

==> (x + 2014)/2010 + (x+2014)/2011 = (x+2014)/2012 + (x+2014)/2013.

==> (x+2014)/2010 + (x+2014)/2011 - (x+2014)/2012 - (x+2014)/2013 = 0

==> (x+2014).(1/2010 + 1/2011 - 1/2012 - 1/2013) = 0

Vì : 1/2010 + 1/2011 - 1/2012 - 1/2013 khác 0

==> x + 2014 = 0

==> x = - 2014

Vậy x = - 2014.

Chúc pạn hok tốt!!!

Bài 1 : làm tương tự với bài 2;3 nhé

Ta có : \(f\left(0\right)=c=2010;f\left(1\right)=a+b+c=2011\)

\(\Rightarrow f\left(1\right)=a+b=1\)

\(f\left(-1\right)=a-b+c=2012\Rightarrow f\left(-1\right)=a-b=2\)

\(\Rightarrow a+b=1;a-b=2\Rightarrow2a=3\Leftrightarrow a=\dfrac{3}{2};b=\dfrac{3}{2}-2=-\dfrac{1}{2}\)

Vậy \(f\left(-2\right)=4a-2b+c=\dfrac{4.3}{2}-2\left(-\dfrac{1}{2}\right)+2010=6+1+2010=2017\)