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\(a\left(a+b+c\right)=-12\)
\(b\left(a+b+c\right)=18\)
\(c\left(a+b+c\right)=30\)
\(a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=-12+18+30\)
\(\left(a+b+c\right)\left(a+b+c\right)=36\)
\(\left(a+b+c\right)^2=\left(\pm6\right)^2\)
\(a+b+c=\pm6\)
Th1:
\(a+b+c=6\)
\(\left[\begin{array}{nghiempt}a\times6=-12\\b\times6=18\\c\times6=30\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=-\frac{12}{6}\\b=\frac{18}{6}\\c=\frac{30}{6}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=-2\\b=3\\c=5\end{array}\right.\)
Th2:
\(a+b+c=-6\)
\(\left[\begin{array}{nghiempt}a\times\left(-6\right)=-12\\b\times\left(-6\right)=18\\c\times\left(-6\right)=30\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=\frac{-12}{-6}\\b=\frac{18}{-6}\\c=\frac{30}{-6}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=2\\b=-3\\c=-5\end{array}\right.\)
\(\frac{a}{2}=\frac{b}{3}\rightarrow\frac{a}{6}=\frac{b}{12};\frac{b}{4}=\frac{c}{5}\rightarrow\frac{b}{12}=\frac{c}{20}\)
Ta có: \(\frac{a}{6}=\frac{b}{12}=\frac{c}{20}\) và a+b-c=3
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU:
\(\frac{a}{6}=\frac{b}{12}=\frac{c}{20}=\frac{a+b-c}{6+12-20}=\frac{3}{-2}\)
*\(\frac{a}{6}=-\frac{3}{2}\rightarrow a=6\cdot-\frac{3}{2}=-9\)
*\(\frac{b}{12}=-\frac{3}{2}\rightarrow b=12\cdot-\frac{3}{2}=-18\)
*\(\frac{c}{20}=-\frac{3}{2}\rightarrow c=20\cdot-\frac{3}{2}=-30\)
\(\Leftrightarrow a+b+c=-8+-18+-30=-56\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)
Khi đó: \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Vậy \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
Ta có abbcca=\(\frac{3}{5}.\frac{4}{5}.\frac{3}{7}\)
=>a2b2c2=\(\frac{36}{175}\)
=>abc=\(\sqrt{\frac{36}{175}}=\frac{6\sqrt{7}}{35}\)
=>a=\(\frac{6\sqrt{7}}{35}:\frac{4}{5}=\frac{3\sqrt{7}}{14}\)=>b=\(\frac{6\sqrt{7}}{35}:\frac{3}{7}=\frac{2\sqrt{7}}{5}\)=>c
