Question

Tính \(Q=a^2+b^2+c^2\)

Biết ac=2; bc=6; ac=3

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Answer 1

Ta có: ab=2 (1)

bc=6 (2)

ac=3 (3)

Nhân (1);(2);(3) :

a.c.b.c.a.c=\(a^2.b^2.c^2=\left(abc\right)^2\)=2.6.3=36

\(\Rightarrow\) abc=6

Mà ab=2

=>c=3

bc=6

=>a=1

ac=3

=>b=2

Answer 2

Ta có:

\(\left\{{}\begin{matrix}ab=2\\bc=6\\ac=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\\dfrac{bc}{ac}=\dfrac{6}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\\dfrac{b}{a}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\b=2a\end{matrix}\right.\)

\(\Leftrightarrow2a^2=2\Leftrightarrow a^2=1\)

Mà:

\(b=2a\Leftrightarrow b^2=4a^2=4\)

\(ac=3\Leftrightarrow a^2c^2=9\Leftrightarrow c^2=9\)

\(\Leftrightarrow Q=a^2+b^2+c^2=1+4+9=14\)

Vậy \(Q=14\)

Answer 3

sai đề rồi