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Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{x^4}{a}+\frac{y^4}{b}\right)(a+b)\geq (x^2+y^2)^2=1\)
\(\Rightarrow \frac{x^4}{a}+\frac{y^4}{b}\geq \frac{1}{a+b}\)
Dấu "=" xảy ra khi \(\frac{x^2}{a}=\frac{y^2}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow \frac{x^{2000}}{a^{1000}}+\frac{y^{2000}}{b^{1000}}=\left(\frac{x^2}{a}\right)^{1000}+\left(\frac{y^2}{b}\right)^{1000}\)
\(=\frac{1}{(a+b)^{1000}}+\frac{1}{(a+b)^{1000}}=\frac{2}{(a+b)^{1000}}\)
x2+y2=1
(x2+y2)2=1
x4+y4+2x2y2=1
thay vào bt ta dc
x4/a+y4/b=x4+y4+2x2y2/a+b
x4b/ab+y4a/ab=x4+y4+2x2y2/a+b
x4b+y4a/a+b=x4+y4+2x2y2/a+b
nhân chéo lên rồi rút gọn ta dc
(x2b-y2a)2=0
x2b=y2a
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
\(3,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\left(\frac{1}{x}\right)^2-2.\frac{1}{x}.\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\frac{1}{x^2}-\frac{2}{xy}+\frac{1}{y^2}\right]-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2}{xy}:\left[\frac{y^2-2.xy+x^2}{x^2y^2}\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}.\frac{x^2y^2}{x^2-2xy+y^2}-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2xy}{x^2-2xy+y^2}+\frac{-x^2-y^2}{x^2-2xy-y^2}\)
\(=\frac{2xy-x^2-y^2}{x^2-2xy+y^2}=\frac{-\left(x^2-2xy+y^2\right)}{x^2-2xy+y^2}=-1\)
\(\frac{2011^3+11^3}{2011^3+2000^3}\)
\(=\frac{\left(2011+11\right)\left(2011^2-2011.11+11^2\right)}{\left(2011+2000\right)\left(2011^2-2011.2000+2000^2\right)}\)
\(=\frac{\left(2011+11\right)\left[2011^2-11\left(2011-11\right)\right]}{\left(2011+2000\right)\left[2011^2-2000\left(2011-2000\right)\right]}\)
\(=\frac{\left(2011+11\right)\left(2011^2-11.2000\right)}{\left(2011+2000\right)\left(2011^2-2000.11\right)}\)
\(=\frac{2011+11}{2011+2000}\left(2011^2-11.2000\ne0\right)\)
đpcm
câu 1 là :từ a/x + b/y + c/z =0 suy ra (ayz+bxz+cxy)/xyz =0 suy ra ayz+bxz+cxy=0 (1)
vì x/a + y/b + z/c =1 (gt) suy ra (x/a + y/b + z/c )^2 = 1^2 . suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(xy/ab + yz/bc + xz/ac) =1
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2[(ayz+bxz+cxy)/abc = 1 (2)
Từ (1) và (2) suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =1 (đpcm)
Câu 3:
a: \(G=\dfrac{a^2}{b\left(a+b\right)}-\dfrac{b^2}{a\left(a-b\right)}+\dfrac{-\left(a^2+b^2\right)}{ab}\)
\(=\dfrac{a^3\left(a-b\right)-b^3\left(a+b\right)-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a^4-a^3b-ab^3-b^4-a^4+b^4}{ab\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{-ab\left(a^2+b^2\right)}{ab\left(a-b\right)\left(a+b\right)}=\dfrac{-a^2-b^2}{a^2-b^2}\)
b: \(\dfrac{a}{b}=\dfrac{a+1}{b+5}\)
nên ab+5a=ab+b
=>5a=b
\(G=\dfrac{-a^2-\left(5a\right)^2}{a^2-\left(5a\right)^2}=\dfrac{-a^2-25a^2}{a^2-25a^2}=\dfrac{-26}{-24}=\dfrac{13}{12}\)