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Đặt \(\hept{\begin{cases}a+b=m\\b+c=n\\c+a=p\end{cases}}\)
Xem VT = A
\(\Rightarrow A=m^2+n^2+p^2-mn-np-mp\)
\(2A=\left(m-n\right)^2+\left(n-p\right)^2+\left(p-m\right)^2\)
\(=\left(a+b-b-c\right)^2+\left(b+c-c-a\right)^2+\left(c+a-a-b\right)^2\)
\(=\left(a-c\right)^2+\left(b-a\right)^2+\left(c-b\right)^2\)
\(=a^2-2ac+c^2+b^2-2ab+a^2+c^2-2bc+b^2\)
\(=2\left(a^2+b^2+c^2-2ab-2bc-2ac\right)\)
\(\Rightarrow A=a^2+b^2+c^2-ab-bc-ca\)(đpcm)
Ta có:
\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)
\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)
\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)
\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Đồng thời:
\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự:
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Từ đó:
\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2b+2c+2a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)
\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)
\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)
(x2 - x + 1)2 - 5x(x2 - x + 1) + 4x2
Đặt x2 - x + 1 = a
<=> a2 - 5xa + 4x2 = x2 - 4xa - xa + 4x2
= a(a - 4x) - x(a - 4x) = (a - x)(a - 4x)
= (x2 - x + 1 - x)(x2 - x + 1 - 4x)
= (x2 - 2x + 1)(x2 - 5x + 1) = (x - 1)2(x2 - 5x + 1)
Đặt x2 - x + 1 = y
đthức <=> y2 - 5xy + 4x2
= y2 - xy - 4xy + 4x2
= y( y - x ) - 4x( y - x )
= ( y - x )( y - 4x )
= ( x2 - x + 1 - x )( x2 - x + 1 - 4x )
= ( x2 - 2x + 1 )( x2 - 5x + 1 )
= ( x - 1 )2( x2 - 5x + 1 )
\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)
\(=\left(a^2+4b^2-5\right)^2-4^2\left(ab+1\right)^2\)
\(=\left(a^2+4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)
\(=\left(a^2+4b^2-5\right)^2-\left[4ab+4\right]^2\)
\(=\left(a^2+4b^2-5-4ab-4\right)\left(a^2+4b^2-5+4ab+4\right)\)
\(=\left(a^2+4b^2-4ab-9\right)\left(a^2+4b^2+4ab-1\right)\)
\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)
= \(\left(a^2+4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)
= \(\left(a^2+4b^2-5\right)^2-\left(4ab+4\right)^2\)
= \(\left(a^2+4b^2-5-4ab-4\right)\left(a^2+4b^2-5+4ab+4\right)\)
= \(\left(a^2+4b^2-4ab-9\right)\left(a^2+4b^2+4ab-1\right)\)
= \(\left[\left(a-2b\right)^2-3^2\right]\left[\left(a+2b\right)^2-1^2\right]\)
= \(\left(a-2b-3\right)\left(a-2b+3\right)\left(a+2b-1\right)\left(a+2b+1\right)\)
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=12\\b\left(a+b+c\right)=18\\c\left(a+b+c\right)=30\end{matrix}\right.\)
\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=12+18+30\)
\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=60\)
\(\Rightarrow\left(a+b+c\right)^2=60\)
\(\Rightarrow a+b+c=\pm\sqrt{60}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\sqrt{60}:12=\dfrac{\sqrt{15}}{6}\\b=\sqrt{60}:18=\dfrac{\sqrt{15}}{9}\\c=\sqrt{60}:30=\dfrac{\sqrt{15}}{15}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\sqrt{60}:12=\dfrac{-\sqrt{15}}{6}\\b=-\sqrt{60}:18=\dfrac{-\sqrt{15}}{9}\\c=-\sqrt{60}:30=\dfrac{-\sqrt{15}}{15}\end{matrix}\right.\end{matrix}\right.\)
Các câu sau làm tương tự
b. \(ab=\dfrac{3}{5};bc=\dfrac{4}{5};ac=\dfrac{3}{4}\)
\(\Rightarrow ab\cdot bc\cdot ac=\dfrac{9}{25}\Rightarrow\left(abc\right)^2=\dfrac{9}{25}\Rightarrow abc=\pm\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\dfrac{3}{5}:bc=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\\b=\dfrac{3}{5}:ac=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5}\\c=\dfrac{3}{5}:ab=\dfrac{3}{5}:\dfrac{3}{5}=1\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\dfrac{3}{5}:\dfrac{4}{5}=-\dfrac{3}{4}\\b=-\dfrac{3}{5}:\dfrac{3}{4}=-\dfrac{4}{5}\\c=-\dfrac{3}{5}:\dfrac{3}{5}=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy......................
