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Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng
Nguyễn Huy Tú Lightning Farron Akai Haruma
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$f(-1)=a-b+c$
$f(-4)=16a-4b+c$
$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$
$\Rightarrow f(-4)=6f(-1)$
$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)
b.
$f(-2)=4a-2b+c$
$f(3)=9a+3b+c$
$\Rightarrow f(-2)+f(3)=13a+b+2c=0$
$\Rightarrow f(-2)=-f(3)$
$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)
a.
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⇒f(−4)=6f(−1)
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⇒f(−1)f(−4)=f(−1).6f(−1)=6[f(−1)]
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b.
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⇒f(−2)=−f(3)
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⇒f(−2)f(3)=−[f(3)]
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn thay mấy giá trị :a,x,b,y đấy vào biểu thức song bấm máy là ra mà.
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(1)
\(\left(ax+by\right)^2+\left(ay-bx\right)^2\)
\(=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)
\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(2)
Từ (1) và (2) ta có \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay-bx\right)^2\)( đpcm )
\(\left(a^2+b^2\right)+\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(\left(ax+by\right)^2+\left(ay-bx\right)^2=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)
\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
Suy ra : \(\left(a^2+b^2\right)+\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay+bx\right)^2\left(đpcm\right)\)