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)

Với x, y khác 0

Ta có: 

\(a^2+b^2=1\Leftrightarrow\left(a^2+b^2\right)^2=1\Leftrightarrow a^4+2a^2b^2+b^4=1\)

Từ bài ra ta suy ra:

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

<=> \(a^4\left(x+y\right)y+b^4\left(x+y\right)x=a^4xy+2a^2b^2xy+b^4xy\)

<=> \(a^4y^2+b^4x^2-2a^2y.b^2x=0\)

<=> \(\left(a^2y-b^2x\right)^2=0\)

<=> \(a^2y-b^2x=0\)

<=> \(a^2y=b^2x\)

Câu b em xem lại đề nhé: Thử \(a=b=\frac{1}{\sqrt{2}};x=y=1\)vào ko thỏa mãn

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

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: \(\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=2\frac{x^{2008}}{a^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

Đặt  \(u=\frac{x}{a};\)  và  \(v=\frac{y}{b}\)  \(\Rightarrow\)  \(\hept{\begin{cases}u,v\in Z\\u+v=1\\uv=-2\end{cases}}\)

Khi đó, ta có:

\(u+v=1\)

nên  \(\left(u+v\right)^3=1\)  \(\Leftrightarrow\)  \(u^3+v^3+3uv\left(u+v\right)=1\)

Do đó,  \(u^3+v^3=1-3uv\left(u+v\right)=1+6=7\)

Vậy,  \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=7\)

\(ĐK:\)  \(a,b,c\ne0\)

Ta có: 

\(a+b+c=0\)

\(\Leftrightarrow\) \(a+b=-c\)

\(\Rightarrow\)  \(\left(a+b\right)^2=\left(-c\right)^2\)

\(\Leftrightarrow\)  \(a^2+b^2+2ab=c^2\)

nên    \(a^2+b^2-c^2=-2ab\)

Tương tự với vòng hoán vị  \(b\rightarrow c\rightarrow a\)  ta cũng suy ra được:

\(\hept{\begin{cases}b^2+c^2-a^2=-2bc\\c^2+a^2-b^2=-2ca\end{cases}}\)

Khi đó, biểu thức  \(P\)  được viết lại dưới dạng:

\(P=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=-\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=-\frac{1}{2}\left(\frac{a+b+c}{abc}\right)=0\) (do \(a,b,c\ne0\)  )

1. Ta có: \(\frac{x}{a}+\frac{y}{b}=1\)

\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}\right)^3=1\)

\(\Rightarrow\left(\frac{x}{a}\right)^3+3.\frac{x}{a}.\frac{y}{b}\left(\frac{x}{a}+\frac{y}{b}\right)+\left(\frac{y}{b}\right)^3=1\)

\(\Rightarrow\left(\frac{x}{a}\right)^3+\left(\frac{y}{b}\right)^3+3.\left(-2\right).1=1\)

\(\Rightarrow\left(\frac{x}{a}\right)^3+\left(\frac{y}{b}\right)^3=1+6=7\)

2.Do \(a+b+c=0\)

Ta có:

\(b^2+c^2-a^2=b^2+c^2+2bc-a^2-2bc\)

                         \(=\left(b+c\right)^2-a^2-2bc\)

                         \(=\left(a+b+c\right)\left(b+c-a\right)-2bc=-2bc\)

CM tương tự: \(a^2+b^2-c^2=-2ab\)

                    \(c^2+a^2-b^2=-2ca\)

Vậy \(P=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)

3.

a)Ta có : \(x^2+y^2=1\Rightarrow x^4+2x^2y^2+y^4=1\Rightarrow x^4+y^4=1-2x^2y^2\)

Ta có :

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

\(\Leftrightarrow\frac{bx^4+ay^4}{ab}=\frac{1}{a+b}\)

\(\Leftrightarrow\left(a+b\right)\left(bx^4+ay^4\right)=ab\)

\(\Leftrightarrow\left(a+b\right)\left(bx^4+ay^4\right)-ab=0\)

\(\Leftrightarrow abx^4+a^2y^4+b^2x^4+aby^4-ab=0\)

\(\Leftrightarrow\left(ay^2\right)^2+\left(bx^2\right)^2+ab\left(x^4+y^4\right)-ab=0\)

\(\Leftrightarrow\left(ay^2\right)^2+\left(bx^2\right)^2+ab-2abx^2y^2-ab=0\)(Do \(x^4+y^4=1-2x^2y^2\))

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

\(\Leftrightarrow ay^2=bx^2\)

b) Ta có : \(x^2+y^2=1\Rightarrow-x^2=y^2-1\)

Xét \(ay^2\left(a+b\right)-ab\)

\(\Leftrightarrow\left(ay\right)^2+aby^2-ab\)

\(\Leftrightarrow\left(ay\right)^2-abx^2\)

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

\(\Rightarrow ay^2\left(a+b\right)-ab=0\)

\(\Rightarrow ay^2\left(a+b\right)=ab\)

\(\Rightarrow\frac{ay^2}{ab}=\frac{1}{a+b}\)

\(\Rightarrow\frac{\left(ay^2\right)^{1004}}{\left(ab\right)^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)

\(\Rightarrow\frac{\left(ay^2\right)^{1004}+\left(bx^2\right)^{1004}}{\left(ab\right)^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

1)

Ta có : a^3+b^3+c^3=(a+b+c).(a^2+b^2+c^2-a.b-b.c-a.c)+3.a.b.c=3.a.b.c

=(a+b+c).(a^2+b^2+c^2-a.b-b.c-a.c)=0

Ta thấy:a,b,c là số dương nên a+b+c khác 0 suy ra (a^2+b^2+c^2-a.b-b.c-a.c) =0 nên a=b=c

Vậy a=b=c

Bài 2:

Từ $xyz=1$ suy ra:

\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=yz+xz+xy\)

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

\(\Leftrightarrow (xy-x-y+1)+yz+xz-z-1=0\)

\(\Leftrightarrow (x-1)(y-1)+yz+xz-z-xyz=0\)

\(\Leftrightarrow (x-1)(y-1)+z(y-1)-xz(y-1)=0\)

\(\Leftrightarrow (y-1)(x-1+z-xz)=0\)

\(\Leftrightarrow (y-1)[(x-1)-z(x-1)]=0\Leftrightarrow (y-1)(x-1)(1-z)=0\)

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

Nếu $x=1\Rightarrow yz=1$

$A=x^{2018}+2019^y-z^x=1+2019^y-z=1+2019^y-\frac{1}{y}$

Nếu $y=1\Rightarrow xz=1$

$A=x^{2018}+2019-z^x=x^{2018}+2019-\frac{1}{x^x}$

Nếu $z=1\Rightarrow xy=1$

$A=\frac{1}{y^{2018}}+2019^y-1$

Tóm lại với đkđb vẫn chưa tính được giá trị cụ thể của $A$

Bài 1:

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow (a+b)^3-3ab(a+b)+c^3=3abc\)

\(\Leftrightarrow (a+b)^3+c^3-3ab(a+b+c)=0\)

\(\Leftrightarrow (a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0\)

\(\Leftrightarrow (a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0\)

\(\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\)

Vì $a,b,c$ dương nên $a+b+c\neq 0$

Do đó $a^2+b^2+c^2-ab-bc-ac=0$

$\Leftrightarrow \frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}=0$

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

Do $(a-b)^2; (b-c)^2; (c-a)^2\geq 0$ với mọi $a,b,c>0$

Suy ra để tổng của chúng bằng $0$ thì $(a-b)^2=(b-c)^2=(c-a)^2=0$

$\Rightarow a=b=c$ (đpcm)

a/ \(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

b/ \(\frac{a}{a+b^2}=\frac{a}{a\left(a+b+c\right)+b^2}=\frac{a}{a^2+b^2+a\left(b+c\right)}\le\frac{a}{2ab+a\left(b+c\right)}=\frac{1}{b+b+b+c}\)

\(\Rightarrow\frac{a}{a+b^2}=\frac{1}{b+b+b+c}\le\frac{1}{16}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right)\)

Tương tự: \(\frac{b}{b+c^2}\le\frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\) ; \(\frac{c}{c+a^2}\le\frac{1}{16}\left(\frac{3}{a}+\frac{1}{c}\right)\)

Cộng vế với vế:

\(VT\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)