bn có thể kb với mk đc ko

Trần Thị Hảo

\(\left\{{}\begin{matrix}a\left(a+b+c\right)=12\\b\left(a+b+c\right)=18\\c\left(a+b+c\right)=30\end{matrix}\right.\)

\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=12+18+30\)

\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=60\)

\(\Rightarrow\left(a+b+c\right)^2=60\)

\(\Rightarrow a+b+c=\pm\sqrt{60}\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\sqrt{60}:12=\dfrac{\sqrt{15}}{6}\\b=\sqrt{60}:18=\dfrac{\sqrt{15}}{9}\\c=\sqrt{60}:30=\dfrac{\sqrt{15}}{15}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\sqrt{60}:12=\dfrac{-\sqrt{15}}{6}\\b=-\sqrt{60}:18=\dfrac{-\sqrt{15}}{9}\\c=-\sqrt{60}:30=\dfrac{-\sqrt{15}}{15}\end{matrix}\right.\end{matrix}\right.\)

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......................

tại sao câu a bn vt phân số \(\frac{-1}{2}\)như rứa vậy :v

Ta có: a (a + b + c) = -12 (1)

         b(a + b+  c) = 18 (2)

       c(a + b + c) = 30 (3)

Từ (1); (2); (3) cộng vế cho vế

=> a(a + b + c) + b(a + b+ c) + c(a +b + c) = -12 + 18 + 30

=> (a + b + c)(a + b + c) = 36

=> (a + b + c)² = 6²

=> \(\orbr{\begin{cases}a+b+c=6\\a+b+c=-6\end{cases}}\)

Thay \(a+b+c=\pm6\) vào (1) ; (2) ;(3):

+) a(a + b + c) = -12 => \(\orbr{\begin{cases}a.6=-12\\a.\left(-6\right)=-12\end{cases}}\) => \(\orbr{\begin{cases}a=-12:6=-2\\a=-12:\left(-6\right)=2\end{cases}}\)

+) b(a + b + c) = 18 => \(\orbr{\begin{cases}b.6=18\\b.\left(-6\right)=18\end{cases}}\) => \(\orbr{\begin{cases}b=18:6=3\\b=18:\left(-6\right)=-3\end{cases}}\)

+) c(a + b+ c) = 30 => \(\orbr{\begin{cases}c.6=30\\c.\left(-6\right)=30\end{cases}}\) => \(\orbr{\begin{cases}c=30:6=5\\c=30:\left(-6\right)=-5\end{cases}}\)

Vậy ...

Ta có: a(a + b + c) = -12

          b(a + b + c) = 18

          c(a + b + c) = 30

=> a(a + b + c) + b(a + b + c) + c(a + b + c) = -12 + 18 + 30

=> (a + b + c)² = 36

=>\(\orbr{\begin{cases}a+b+c=6\\a+b+c=-6\end{cases}}\)=> \(\orbr{\begin{cases}a=-2;b=3;c=5\\a=2;b=-3;c=-5\end{cases}}\)

Vậy: a = 2; -2

        b = 3; -3

        c = 5; -5