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c.
\(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=2010\)
\(\leftrightarrow\) \(x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+1+x^2+y^2+x^2y^2=2010\)
\(\leftrightarrow\)\(x^2+x^2y^2+2x\sqrt{1+y^2}.y\sqrt{1+x^2}+y^2+x^2y^2=2009\)
\(\leftrightarrow\) \(\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=2009\)
\(\leftrightarrow\) \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=\sqrt{2009}\)
c) \(A^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2y^2+x^2+x^2y^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2-1=2010-1=2009\)
Vì A>0 nên \(A=\sqrt{2009}\)
d) \(2009^2=\left(2008+1\right)^2=2008^2+2.2008+1\)
\(1+2008^2=2009^2-2.2008=2009^2-2.2009\dfrac{2008}{2009}\)
\(A=\sqrt{2009^2-2.2009.\dfrac{2008}{2009}+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}\)
\(A=\sqrt{\left(2009-\dfrac{2008}{2009}\right)^2}+\dfrac{2008}{2009}=2009-\dfrac{2008}{2009}+\dfrac{2008}{2009}=2009\)
Lời giải:
Đặt \(\sqrt{x-1}=a; \sqrt{y-1}=b(a,b\geq 0)\)
\(\Rightarrow x=a^2+1; y=b^2+1\). PT trở thành:
\(\frac{(a^2+1)^2-4}{a^2+1}+\frac{(b^2+1)^2-4}{b^2+1}+8=4(a+b)\)
\(\Leftrightarrow \frac{(a^2+1)^2-4}{a^2+1}+4-4a+\frac{(b^2+1)^2-4}{b^2+1}+4-4b=0\)
\(\Leftrightarrow a^2+5-4a-\frac{4}{a^2+1}+b^2+5-4b-\frac{4}{b^2+1}=0\)
\(\Leftrightarrow (a^2-4a+3)+2-\frac{4}{a^2+1}+(b^2-4b+3)+2-\frac{4}{b^2+1}=0\)
\(\Leftrightarrow (a-1)(a-3)+\frac{2(a^2-1)}{a^2+1}+(b-1)(b-3)+\frac{2(b^2-1)}{b^2+1}=0\)
\(\Leftrightarrow (a-1)\left(a-3+\frac{2(a+1)}{a^2+1}\right)+(b-1)\left(b-3+\frac{2(b+1)}{b^2+1}\right)=0\)
\(\Leftrightarrow (a-1).\frac{a^3-3a^2+3a-1}{a^2+1}+(b-1).\frac{b^3-3b^2+3b-1}{b^2+1}=0\)
\(\Leftrightarrow \frac{(a-1)(a-1)^3}{a^2+1}+\frac{(b-1)(b-1)^3}{b^2+1}=0\)
\(\Leftrightarrow \frac{(a-1)^4}{a^2+1}+\frac{(b-1)^4}{b^2+1}=0\)
Dễ thấy mỗi số hạng ở vế trái đều không âm. Do đó để tổng của chúng bằng $0$ thì \(\frac{(a-1)^4}{a^2+1}=\frac{(b-1)^4}{b^2+1}=0\Rightarrow a=b=1\Rightarrow x=y=2\)
Vậy.........
Đặt \(A = \frac{x^4 - 1}{y + 1} + \frac{y^4 - 4}{x + 1}\)
\(=\frac{(x^4 - 1)(x + 1) + (y^4 - 4)(y + 1)}{(y + 1)(x + 1)}\)
=>\(A(y + 1) = \frac{(x^4 - 1)(x + 1) + (y^4 - 4)(y + 1)}{x + 1} = \frac{(x^4 - 1)(x + 1)}{x + 1} + \frac{(y^4 - 4)(y + 1)}{x + 1}\)
=>\(A(y + 1) = (x^4 - 1) + \frac{(y^4 - 4)(y + 1)}{x + 1}\)
A∈Z
=>A(y+1)∈Z
=>\((y^4 - 4)(y + 1) \vdots (x + 1)\)
Đặt \((y^4-4)(y+1)=k\left(x+1)\left(\right.k\in\mathbb{Z}\right)\)
=>\(A = \frac{x^4 - 1}{y + 1} + \frac{k(x + 1)}{x + 1} = \frac{x^4 - 1}{y + 1} + k\)
A∈Z và k∈Z
=>\((x^4 - 1) \vdots (y + 1) \quad (1)\)
\(x^4y^{44} - 1 = x^4y^{44} - y^{44} + y^{44} - 1 = y^{44}(x^4 - 1) + (y^{44} - 1)\)
(1)=>\(y^{44}(x^4-1)\vdots(y+1)\)
Ta có: \(y^{44} - 1 = (y^2)^{22} - 1^2 = (y^2 - 1)(y^{42} + y^{40} + \dots + 1)\)
mà \(y^2 - 1 = (y - 1)(y + 1) \vdots (y + 1)\)
nên \((y^{44}-1)\vdots(y+1)\)
Do đó: \(y^{44}(x^4 - 1) + (y^{44} - 1) \vdots (y + 1)\)
=>\(x^4y^{44}-1\vdots(y+1)\quad\) (ĐPCM)