Question

[Dm tam]

Cho a+b=m ; a-b=n. Tính ab và a³-b³ theo m và n

Answer 1

[T.Ps]

#)Giải :

Ta có : 

\(\left(a+b\right)\left(a-b\right)=a^2-b^2=m.n\)

\(\left(a+b\right)\left(a^2+b^2\right)=a^3-ab^2+a^2b-b^3=m.n.m=m^2n\)

Lại có :

\(a^2+2ab+b^2=m^2\)

\(a^2-2ab+b^2=n^2\)

\(\Rightarrow4ab=m^2-n^2\)

\(\Rightarrow ab=\frac{m^2-n^2}{4}\)

\(\Rightarrow a^3-b^3=m^2n-\frac{m^2-n^2}{4}n\)

Answer 2

[ღᏠᎮღ🅼🅸🅽🅷ঔ🅽🅶ụ🆈ঌ__[ᴅʀᴇᴀм тᴇᴀм]__]

Vì \(\hept{\begin{cases}a+b=m\\a-b=n\end{cases}\Rightarrow\hept{\begin{cases}a=m-b\\m-b-b=n\end{cases}\Leftrightarrow}\hept{\begin{cases}a=m-b\\m-2b=n\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}a=m-b\\b=-\frac{\left(n-m\right)}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}a=m-b\\b=\frac{m-n}{2}\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}a=m-\frac{m-n}{2}\\b=\frac{m-n}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{2m-m+n}{2}\\b=\frac{m-n}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=\frac{m+n}{2}\\b=\frac{m-n}{2}\end{cases}}\)

Do đó \(a.b=\frac{m+n}{2}.\frac{m-n}{2}=\frac{m^2-n^2}{4}\)

+) \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

                    \(=n.\left[\left(\frac{m+n}{2}\right)^2+\frac{m^2-n^2}{4}+\left(\frac{m-n}{2}\right)^2\right]\)

                    \(=n.\frac{m^2+2mn+n^2+m^2-n^2+m^2-2mn+n^2}{4}\)

                    \(=n.\frac{3m^2-n^2}{4}\)

                  \(=\frac{3m^2n\cdot n^3}{4}\)

_Minh ngụy_