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Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{x^4}{a}+\frac{y^4}{b}\right)(a+b)\geq (x^2+y^2)^2=1\)
\(\Leftrightarrow \frac{x^4}{a}+\frac{y^4}{b}\geq \frac{1}{a+b}\)
Dấu bằng xảy ra khi \(\frac{x^2}{a}=\frac{y^2}{b}\). Do đó \(\frac{x^2}{a}=\frac{y^2}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow \frac{x^{2006}}{a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{(a+b)^{1003}}\)
\(\Rightarrow \frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{y^{1003}}=\frac{2}{(a+b)^{1003}}\)
Do đó ta có đpcm.
Bài này phải quy đồng rồi áp dụng chớ chớ lỡ a+b=0 thì sao chị :3
\(\text{Đặt }x^2=m\ge0;y^2=n\ge0\Rightarrow m+n=1\)
\(\text{Ta có: }\frac{m^2}{a}+\frac{n^2}{b}=\frac{\left(m+n\right)^2}{a+b}\Leftrightarrow\left(a+b\right)\left(\frac{m^2}{a}+\frac{n^2}{b}\right)=\left(m+n\right)^2\left(\text{BĐT Bunhiacopki}\right)\)\(\Leftrightarrow m^2+n^2+\frac{b}{a}m^2+\frac{a}{b}n^2=m^2+n^2+2mn\)
\(\Leftrightarrow\frac{b}{a}m^2+\frac{a}{b}n^2-2mn=0\left(1\right)\)
\(\text{+Nếu }\frac{a}{b}< 0\text{ thì (1)}\Leftrightarrow-\left(\sqrt{-\frac{b}{a}m}\right)^2-2mn-\left(\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}\right)^2=0\)
\(\Leftrightarrow\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}=0\Leftrightarrow m=n=0\left(\text{loại}\right)\)
\(\text{Xét }\frac{a}{b}>0;\left(1\right)\Leftrightarrow\left(\sqrt{\frac{b}{a}m}\right)^2-2mn+\left(\sqrt{\frac{a}{b}n}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}-\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\sqrt{\frac{b}{a}m}=\sqrt{\frac{a}{b}n}\)
\(\Leftrightarrow bm=an\Leftrightarrow bx^2=ay^2\left(a,b>0\right)\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\left(\frac{x^2}{a}\right)^{1003}+\left(\frac{y^2}{b}\right)^{1003}=\frac{1}{\left(a+b\right)^{1003}}+\frac{1}{\left(a+b\right)^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\left(đpcm\right)\)
1/ Ta có: \(\frac{x^4}{1a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow1bx^4\left(a+b\right)+ay^4\left(a+b\right)=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow\left(ay^2-bx^2\right)^2=0\)
\(\Rightarrow\frac{x^2}{1a}=\frac{y^2}{b}=\frac{\left(x^2+y^2\right)}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2006}}{1a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{\left(a+b\right)^{1003}}\)
\(\Rightarrow\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
