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21 tháng 3 2018

\(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2001}-\frac{1}{2002}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)

=  \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)

\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)

4 tháng 12 2018

tui mới học lớp 6 thui

3 tháng 11 2019

Ta có:

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

Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:

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

\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)

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

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

\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)

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

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

\(\Leftrightarrow x^2b=y^2a\)

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

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)

Chúc bạn học tốt!

14 tháng 4 2017

TUI MỚI LỚP 5 THÔI

23 tháng 1 2016

6567 đồng

tick nha

14 tháng 4 2017

@@Ace Legona giúp mk với

14 tháng 8 2020

.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)

.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)

.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)

.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)

.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)

.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)

.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)

.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)

.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)

.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)

.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)

.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)

.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)

.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)

.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)

.Suy ra \(x-1000=0\Leftrightarrow x=1000\)

14 tháng 8 2020

cảm ơn

26 tháng 2 2019

Ta có : 

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) vì \(x^2+y^2=1\)

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

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

\(\Leftrightarrow\left(x^4.b+y^4.a\right)\left(a+b\right)=ab\left(x^2+y^2\right)^2\)

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

\(=ab\left(x^2+y^2\right)\left(x^2+y^2\right)\)

\(\Rightarrow ab\left(x^4+x^2y^2+x^2y^2+y^4\right)\)

\(\Rightarrow abx^4+abx^2y^2+abx^2y^2+abx^2y^2+aby^4\)

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

\(=2abx^2y^2\)

\(\Rightarrow\left(bx^2\right)^2+\left(ay^2\right)^2-ax^2.by^2-ax^2-by^2=0\)

\(\Rightarrow\left[\left(bx^2\right)^2-ax^2.by^2\right]+\left[\left(ay^2\right)^2-ax^2.by^2\right]=0\)

\(bx^2\left(bx^2-ay^2\right)+ay^2\left(ay^2-bx^2\right)=0\)

\(bx^2\left(bx^2-ay^2\right)-ay^2\left(bx^2-ay^2\right)\)

\(\left(bx^2-ay^2\right)^2=0\)

\(bx^2-ay^2=0\)

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

Mà \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Rightarrow x^2.\frac{x^2}{a}+y.\frac{y^2}{b}=\frac{x^2+y^2}{a+b}\)

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

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

Ta có :

\(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{a^{1002}}=\left(\frac{x^2}{a}\right)^{1002}+\left(\frac{y^2}{b}\right)^{1002}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}< đpcm>\)

Hok tốt 

P/s : _Làm bừa nên chắc k đúng đâu - - _M bt a hok ngu thek nào r mak (:

26 tháng 2 2019

_E cóa thý a hok ngu âu >: ?

_Với cả giải vợi lak đầy đủ roy hả ?

_Thank nhìu nhìu <<<: