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7 tháng 7 2016

a)Ta có

\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)

\(\Rightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Rightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+y^4+2x^2y^2}{a+b}\)

\(\Rightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+y^2-2x^2y^2\right)ab\)

\(\Rightarrow x^4ab+x^4b^2+y^4ab+y^4a^2=x^4ab+y^4ab+2x^2y^2ab\)

\(\Rightarrow x^4b^2+y^4b^2-2x^2y^2ab=0\)

\(\Rightarrow\left(x^2b-y^2a\right)^2=0\)

\(\Rightarrow x^2b-y^2a=0\)

\(\Rightarrow x^2b=y^2a\left(dpcm\right)\)

b) từ kết quả câu a) ta suy ra dc

\(\frac{x^2}{a}=\frac{y^2}{b}\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}\)

Mà \(x^2+y^2=1\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{1}{a+b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1005}=\left(\frac{y^2}{b}\right)^{1005}=\frac{1^{1005}}{\left(a+b\right)^{1005}}\Rightarrow\frac{x^{2010}}{a^{1005}}=\frac{y^{2010}}{b^{1005}}=\frac{1}{\left(a+b\right)^{1005}}\)

\(\Rightarrow\frac{x^{2010}}{a^{1005}}+\frac{y^{2010}}{b^{1005}}=\frac{1}{\left(a+b\right)^{1005}}+\frac{1}{\left(a+b\right)^{1005}}=\frac{2}{\left(a+b\right)^{1005}}\left(dpcm\right)\)

Vầy đúng không nhỉ nếu đúng T I C K cho mình nha 

Ko biết có nhanh nhất ko nhưng dù sao cũng xong rồi

8 tháng 9 2017

https://olm.vn/hoi-dap/question/1038454.html 

Mình vừa làm cách đây 11 phút nhé !

Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005 

<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005 

<=> 2a2010 + 2b2010 + 2c2010  - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0

<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005  +  a2010​) = 0

<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 ​)2 = 0

=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 ​ = 0

=> a = b = c 

Vậy (a - b)20 + (b - c)11 + (c - a)2010 = (a - a)20 + (a - a)11 + (a - a)2010 = 0 + 0 + 0 = 0 .

8 tháng 9 2017

Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005 

<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005 

<=> 2a2010 + 2b2010 + 2c2010  - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0

<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005  +  a2010​) = 0

<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 ​)2 = 0

=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 ​ = 0

=> a = b = c 

Vậy (a - b)20 + (b - c)11 + (c - a)2010 = (a - a)20 + (a - a)11 + (a - a)2010 = 0 + 0 + 0 = 0 .

19 tháng 2 2018

       a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005 

<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005 

<=> 2a2010 + 2b2010 + 2c2010  - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0

<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005  +  a2010​) = 0

<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 ​)2 = 0

=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 ​ = 0

=> a = b = c 

NV
3 tháng 4 2019

Đặt \(\left\{{}\begin{matrix}a^{1005}=x\\b^{1005}=y\\c^{1005}=z\end{matrix}\right.\) \(\Rightarrow x^2+y^2+z^2=xz+xz+yz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2xz+2yz\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-z=0\\y-z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z\)

\(\Rightarrow a^{1005}=b^{1005}=c^{1005}\Rightarrow a=b=c\)

\(\Rightarrow M=0\)

9 tháng 6 2019

bạn cũng xem phim gia sư siêu quậy reborn à ?

18 tháng 5 2016

\(\Leftrightarrow2\left(a^{2010}+b^{2010}+c^{2010}\right)=2\left(a^{1005}b^{1005}+b^{1005}c^{1005}+c^{1005}a^{1005}\right)\)

\(\Leftrightarrow2a^{2010}+2b^{2010}+2c^{2010}-2a^{1005}b^{1005}-2b^{1005}c^{1005}-2c^{1005}a^{1005}=0\)

\(\Leftrightarrow\left(a^{2010}-2a^{1005}b^{1005}+b^{2010}\right)+\left(b^{2010}-2b^{1005}c^{1005}+c^{2010}\right)+\left(c^{2010}-2c^{1005}a^{1005}+a^{2010}\right)=0\)

\(\Leftrightarrow\left(a^{1005}-b^{1005}\right)^2+\left(b^{1005}-c^{1005}\right)^2+\left(c^{1005}-a^{1005}\right)^2=0\)

\(\Rightarrow\left(a^{1005}-b^{1005}\right)^2=0;\left(b^{1005}-c^{1005}\right)^2=0;\left(c^{1005}-a^{1005}\right)^2=0\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\left(a-a\right)^{20}+\left(a-a\right)^{11}+\left(a-a\right)^{2010}=0\)

2 ( a trên 2010 + b trân 2010 + c trên 2010 ) = 2 ( a trên 1005 b trên 1005 + b trên 1005 c trên 1005 + c trên 1005 a trên 1005 )

2a^ ( 2010 ) + 2b^ ( 2010 ) + 2c^ ( 2010 ) - 2a^ ( 1005 ) b^ ( 1005 ) - 2b^ ( 1005 ) c^ ( 1005 ) - 2c^ ( 1005 )a^ ( 1005 ) = O\)

( a^ ( 2010 ) - 2a^ ( 1005 ) b^ ( 1005 ) + b^ ( 2010 ) + ( b^( 2010 ) - 2b^ ( 1005 ) c^ ( 1005 ) + c^ ( 2010 ) + ( c^ ( 2010 ) - 2c^ ( 1005 ) a^ ( 1005 ) + a^ ( 2010 ) = 0\)

( a^ ( 1005 ) ^2 + ( b^ ( 1005 ) - c^ ( 1005 ) ^2 + ( c^ ( 1005 ) - a^ ( 1005 ) - a^ ( 1005 ) ^2 = 0\)

( a^ ( 1005 ) - b^ ( 1005 ) ^ 2= 0 : ( b^ ( 1005 ) - c^ ( 1005 ) ^2 = 0 : ( c^ ( 1005 ) - a^ ( 1005 ) ^2 = 0\)

a = b = c

( a - a ) ^ ( 20 ) + ( a - a ) ^ ( 11 ) + ( a - a ) ^ (2010 = 0\)

Vậy :  ( a -a ) ^ ( 20 ) + ( a - a ) ^ ( 11 ) + ( a + a ) ^ ( 2010 = 0\)

\(a^{2010}+b^{2010}+c^{2010}=a^{1005}b^{1005}+b^{1005}c^{1005}+a^{1005}c^{1005}\)

=>\(2a^{2010}+2b^{2010}+2c^{2010}-2a^{1005}b^{1005}-2b^{1005}c^{1005}-2a^{1005}c^{1005=0}\)

=>\(\left(a^{1005}-b^{1005}\right)\left(b^{1005}-c^{1005}\right)\left(a^{1005}-c^{1005}\right)=0\)

=>a=b=c

\(A=\left(b-b\right)^{20}+\left(b-b\right)^{11}+\left(c-c\right)^{2010}=0\)