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Ta có x^4/a + y^4/b = 1/(a + b)
<=> x^4/a + y^4/b = (x^2 + y^2)^2/(a + b).
Bn tự qui đồng và khử mẫu nha, xong thì đc : (a + b)(bx^4 + ay^4) = ab(x^4 + 2x^2y^2 + y^4)
<=> abx^4 + a^2y^4 + b^2x^4 + aby^4 = abx^4 + 2abx^2y^2 + aby^4
<=> a^2y^4 - 2abx^2y^4 + b^2x^4 = 0
<=> (ay^2 - bx^2)^2 = 0
<=> ay^2 - bx^2 = 0
<=> bx^2 = ay^2 => đpcm
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a) Từ đề bài \(\Rightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\) \(\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)-ab\left(x^2+y^2\right)^2=0\)
\(\Leftrightarrow b^2x^4-2abx^2y^2+a^2y^4=0\)
\(\Leftrightarrow\left(bx^2-ay^2\right)^2=0\) \(\Rightarrow bx^2=ay^2\) (ĐPCM)
b) Từ a \(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}\) Áp dụng DTSBN ta có :
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}\) hay \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2018}}{a^{1004}}=\frac{y^{2018}}{b^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\) \(\Rightarrow\frac{x^{2018}}{a^{1004}}+\frac{y^{2018}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\) (ĐPCM)
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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
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Em vào câu hỏi tương tự tham khảo:
a) Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)
Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)
<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)
<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)
<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)
a) \(\frac{x^2}{a}=\frac{y^2}{b}\Leftrightarrow bx^2=ay^2\)
b) \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)
Khi đó: \(\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=2\frac{x^{2008}}{a^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)
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a) \(xy+1-x-y\)
\(=x\left(y-1\right)-\left(y-1\right)\)
\(=\left(y-1\right)\left(x-1\right)\)
b) \(ax+ay-3x-3y\)
\(=a\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(a-3\right)\)
c) \(x^3-2x^2+2x-4\)
\(=x^2\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2\right)\)
d) \(x^2+ab+ax+bx\)
\(=x\left(b+x\right)+a\left(b+x\right)\)
\(=\left(b+x\right)\left(a+x\right)\)
e) \(16-x^2+2xy-y^2\)
\(=16-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
f) \(ax^2+ax-bx^2-bx-a+b\)
\(=\left(ax^2+ax-a\right)-\left(bx^2+bx-b\right)\)
\(=a\left(x^2+x-1\right)-b\left(x^2+x-1\right)\)
\(=\left(x^2+x-1\right)\left(a-b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) pp: biến đổi tương đương
ta có: VT= \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+x^2\right).\)
= \(\left(ax\right)^2+\left(ay\right)^2+\left(az\right)^2+\left(bx\right)^2+\left(by\right)^2+\left(bz\right)^2+\left(cx\right)^2+\left(cy\right)^2+\left(cz\right)^2\) (*)
VP=\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)+\left(bz-cy\right)^2+\left(cx-az\right)^2+\left(ay-bx\right)^2\)
=\(\: \left(ax\right)^2+\left(by\right)^2+\left(cz\right)^2+2\left(axby+bycz+czax\right)+\left(bz\right)^2+\left(cy\right)^2+\left(cx\right)^2+\left(az\right)^2\)
\(+\left(ay\right)^2+\left(bx\right)^2-2\left(bzcy+cxaz+aybx\right)\) (**)
Từ (*),(**)=> VT-VP=0=> VT=VP=> \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+x^2\right).\)=\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)+\left(bz-cy\right)^2+\left(cx-az\right)^2+\left(ay-bx\right)^2\) (đpcm)
2) áp dụng BĐT Schwartz ta có:
\(\left(a+b+c\right)^2\le\left(1+1+1\right)\left(a^2+b^2+c^2\right)\)
=>\(2010^2\le3\left(a^2+b^2+c^2\right)\) (vì a+b+c=2010)
=>\(a^2+b^2+c^2\ge\frac{2010^2}{3}=1346700\)
Dấu '=' xảy ra khi: a=b=c
Vậy GTNN của a^2 +b^2 +c^2 là 1346700 khi a=b=c
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\(a,xy+1-x-y\)
\(=\left(xy-y\right)+\left(1-x\right)\)
\(=y\left(x-1\right)- \left(x-1\right)\)
\(=\left(x-1\right)\left(y-1\right)\)
\(b,ax+ay-3x-3y\)
\(=a\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(a-3\right)\)
\(c,x^3-2x^2+2x-4\)
\(=x^2\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x^2+2\right)\left(x-2\right)\)
\(d,x^2+ab+ax+bx\)
\(=\left(x^2+ax\right)+\left(ab+bx\right)\)
\(=x\left(a+x\right)+b\left(a+x\right)\)
\(=\left(a+x\right)\left(b+x\right)\)
\(e,16-x^2+2xy-y^2\)
\(=4^2-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)