\(\left(ax+by\right)^2=1\Leftrightarrow\left(ax\right)^2+2abxy+\left(by\right)^2=1\Leftrightarrow2xy\le1\Leftrightarrow xy\le\frac{1}{2}\)

Đáp án là D

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

\(\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+y^4+2x^2y^2}{a+b}\Leftrightarrow\left(a+b\right)\left(x^4b+y^4a\right)=ab\left(x^4+y^4+2x^2y^2\right)\)

\(\Leftrightarrow x^4ab+y^4a^2+x^4b^2+y^4ab=x^4ab+y^4ab+2x^2y^2ab\)

\(\Leftrightarrow y^4a^2+x^4b^2=2x^2y^2ab\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Leftrightarrow\frac{x^2}{a}=\frac{y^2}{b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1001}=\left(\frac{y^2}{b}\right)^{1001}\Leftrightarrow\frac{x^{2002}}{a^{1001}}=\frac{y^{2002}}{b^{2011}}\)

Mà: \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\Leftrightarrow\left(\frac{x^2}{a}\right)^{1001}=\frac{1}{\left(a+b\right)^{1001}}\)

\(\Rightarrow\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\frac{2}{\left(a+b\right)^{1001}}\left(đpcm\right)\)

\(x^2+y^2=1\Rightarrow y^2=1-x^2\)

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

\(\Leftrightarrow bx^4+ay^4=\frac{ab}{a+b}\Leftrightarrow bx^4+a\left(1-x^2\right)^2-\frac{ab}{a+b}=0\)

\(\Leftrightarrow bx^4+a\left(x^4-2x^2+1\right)-\frac{ab}{a+b}=0\)

\(\Leftrightarrow\left(a+b\right)x^4-2ax^2+a-\frac{ab}{a+b}=0\)

\(\Leftrightarrow\left(a+b\right)x^4-2ax^2+\frac{a^2}{a+b}=0\Leftrightarrow\left(a+b\right)\left[x^4-2.x.\frac{a}{a+b}+\left(\frac{a}{a+b}\right)^2\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(x^2-\frac{a}{a+b}\right)=0\Rightarrow x^2=\frac{a}{a+b}\) (do \(a+b\ne0\))

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

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

Thay vào bài toán:

\(\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\left(\frac{x^2}{a}\right)^{1001}+\left(\frac{y^2}{b}\right)^{1001}=\left(\frac{1}{a+b}\right)^{1001}+\left(\frac{1}{a+b}\right)^{1001}=\frac{2}{\left(a+b\right)^{1001}}\)

$x^2+y^2=1\Rightarrow y^2=1-x^2$

$\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\iff \frac{b.x^4+a.y^4}{ab}=\frac{1}{a+b}$

$\iff b{x}^4+a{y}^4=\frac{ab}{a+b}\iff b{x}^4+a{\left(1-x^2\right)}^2-\frac{ab}{a+b}=0$

$\iff b{x}^4+a\left(x^4-2x^2+1\right)-\frac{ab}{a+b}=0$

$\iff \left(a+b\right)x^4-2a{x}^2+a-\frac{ab}{a+b}=0$

$\iff \left(a+b\right)x^4-2a{x}^2+\frac{a^2}{a+b}=0\iff \left(a+b\right)\left[x^4-2.x.\frac{a}{a+b}+{\left(\frac{a}{a+b}\right)}^2\right]=0$

$\iff \left(a+b\right)\left(x^2-\frac{a}{a+b}\right)=0\Rightarrow x^2=\frac{a}{a+b}$ (do $a+b\ne 0$)

$\Rightarrow y^2=1-x^2=\frac{b}{a+b}$

$\Rightarrow$ $\frac{x^2}{a}=\frac{a}{a\left(a+b\right)}=\frac{1}{a+b}$ ; $\frac{y^2}{b}=\frac{b}{b\left(a+b\right)}=\frac{1}{a+b}$

Thay

$\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}={\left(\frac{x^2}{a}\right)}^{1001}+{\left(\frac{y^2}{b}\right)}^{1001}={\left(\frac{1}{a+b}\right)}^{1001}+{\left(\frac{1}{a+b}\right)}^{1001}=\frac{2}{{\left(a+b\right)}^{1001}}$

Đáp án C

Xét pt tương giao:

2 x - 1 x - 1 = x + m   ⇔ 2 x - 1 - x + m x - 1 = 0 ⇔ x 2 - 3 - m x + m - 1 = 0

a + b 2 - 4 a b = 8 ⇔ 3 - m 2 - 4 1 - m = 8 ⇔ [ m = - 1 m = 3

Đáp án C