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x-1009/1001+x-4/1003+x+2010/1005=7
((x-1009/1001)-1))+((x-4/1003)-2)+((x+2010/1005)-4))=0
(x-2010/1001)+(x-2010/1003)+(x-2010/1005)=0
(x-2010)*(1/1001+1/1003+1/1005)=0
okk!!!!!!!!!!!!!!!
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Rightarrow x=2010\)
Vậy....
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(x-2010=0\)
\(x=2010\)
Vậy x = 2010
\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)
\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)
\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)
\(\Leftrightarrow x-2007=0\)
\(\Leftrightarrow x=2007\)
\(\text{Đặt }x^2=m\ge0;y^2=n\ge0\Rightarrow m+n=1\)
\(\text{Ta có: }\frac{m^2}{a}+\frac{n^2}{b}=\frac{\left(m+n\right)^2}{a+b}\Leftrightarrow\left(a+b\right)\left(\frac{m^2}{a}+\frac{n^2}{b}\right)=\left(m+n\right)^2\left(\text{BĐT Bunhiacopki}\right)\)\(\Leftrightarrow m^2+n^2+\frac{b}{a}m^2+\frac{a}{b}n^2=m^2+n^2+2mn\)
\(\Leftrightarrow\frac{b}{a}m^2+\frac{a}{b}n^2-2mn=0\left(1\right)\)
\(\text{+Nếu }\frac{a}{b}< 0\text{ thì (1)}\Leftrightarrow-\left(\sqrt{-\frac{b}{a}m}\right)^2-2mn-\left(\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}\right)^2=0\)
\(\Leftrightarrow\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}=0\Leftrightarrow m=n=0\left(\text{loại}\right)\)
\(\text{Xét }\frac{a}{b}>0;\left(1\right)\Leftrightarrow\left(\sqrt{\frac{b}{a}m}\right)^2-2mn+\left(\sqrt{\frac{a}{b}n}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}-\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\sqrt{\frac{b}{a}m}=\sqrt{\frac{a}{b}n}\)
\(\Leftrightarrow bm=an\Leftrightarrow bx^2=ay^2\left(a,b>0\right)\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\left(\frac{x^2}{a}\right)^{1003}+\left(\frac{y^2}{b}\right)^{1003}=\frac{1}{\left(a+b\right)^{1003}}+\frac{1}{\left(a+b\right)^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\left(đpcm\right)\)
\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)
\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)
\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)
Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
1/ Ta có: \(\frac{x^4}{1a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow1bx^4\left(a+b\right)+ay^4\left(a+b\right)=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow\left(ay^2-bx^2\right)^2=0\)
\(\Rightarrow\frac{x^2}{1a}=\frac{y^2}{b}=\frac{\left(x^2+y^2\right)}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2006}}{1a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{\left(a+b\right)^{1003}}\)
\(\Rightarrow\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\)
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
\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)=7
⇔\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)-7=0
⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)
⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
⇔(x-2010)\(\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)\)=0
⇔x-2010=0
⇔x=2010
Vậy x=2010
\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
⇔ \(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}-7=0\)
⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)\)\(+\left(\frac{x+2010}{1005}-4\right)=0\)
⇔\(\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\)\(\frac{x+2010-4020}{1005}=0\)
⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
⇔\(\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)
⇔ \(x-2010=0\left(do\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}>0\right)\)
⇔ \(x=2010\)
Vậy S = {2010}
\(\frac{x-1003}{1007}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
\(\Rightarrow\left(\frac{x-1003}{1007}-1\right)+\left(\frac{x-4}{1003}-1\right)+(\frac{x+2010}{1005}-4)=0\)
\(\Rightarrow\frac{x-2010}{1007}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
\(\Rightarrow\left(x-2010\right)\left(\frac{1}{1007}+\frac{1}{1003}+\frac{1}{1005}\right)\)
Vì
\(\frac{1}{1007}+\frac{1}{1003}+\frac{1}{1005}\ne0\Rightarrow X-2010=0\Rightarrow x=2010\)
\(\frac{x-1003}{1007}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
\(\frac{x-1003}{1007}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
\(\frac{x-2010}{1003}+\frac{x-2010}{1005}+\frac{x-2010}{1007}=0\)
\(\left(x-2010\right)\left(\frac{1}{1003}+\frac{1}{1005}+\frac{1}{1007}\right)=0\)
\(\frac{1}{1003}+\frac{1}{1005}+\frac{1}{1007}\ne0\)
\(\Rightarrow x-2010=0\Rightarrow x=2010\)