Question

[Nguyen duc thanh]

a) Chứng mimh rằng nếu: a + b + c + d = 0 thì:

                             a³ + b³ + c³ + d³ = 3( ac - bd ).( b + d ).

b) Tìm các số tự nhiên a, b , c thỏa mãn điều kiện:

           \(\hept{\begin{cases}\sqrt{a-b+c=\sqrt{a}-\sqrt{b}+\sqrt{c}}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\end{cases}}\)

Answer 1

[Nguyen duc thanh]

a) a + b + c + d = 0 \(\Rightarrow a+c=-\left(b+d\right)\)

\(\Rightarrow\)\(\left(a+c\right)^3=-\left(b+d\right)^3\)

\(\Rightarrow\)\(a^3+c^3+3ac\left(a+c\right)=-b^3-d^3-3b\left(b+d\right)\)

\(\Rightarrow\)\(a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\)

                                             \(=3\left(ac-bd\right)\left(b+d\right)\)\(\left(dpcm\right)\)

b) - \(\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\)

\(\Leftrightarrow\left(\sqrt{a-b+c}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sqrt{c}\right)^2\)

\(\Leftrightarrow b\left(a-b+c\right)=ac\Leftrightarrow\left(b-c\right)\left(a-b\right)=0\Leftrightarrow\orbr{\begin{cases}a=b\\b=c\end{cases}\left(1\right)}\)

   - Gia su \(a\le b\le c\), ta có: \(1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{a}\)

\(\Rightarrow a\le3\Rightarrow a=1,2,3\)

      + Nếu a = 1 thì: \(\frac{1}{b}+\frac{1}{c}=0\left(vl\right)\)

      + Nếu a = 2 thì: \(\frac{1}{b}+\frac{1}{c}=\frac{1}{2}\le\frac{2}{b}\Rightarrow b\le4\)

\(\Rightarrow a=2;b=c=4\)

      + Nếu a = 3 thì: \(\frac{1}{b}+\frac{1}{c}=\frac{2}{3}\le\frac{2}{b}\Rightarrow b\le3\)

\(\Rightarrow a=b=c=3\)

Cac cap (a, b, c) thoa \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)la:

        \(\left(2,4,4\right);\left(4,2,4\right);\left(4,4,2\right);\left(3,3,3\right)\)

Kết hợp với \(\left(1\right)\)ta có nghiệm: \(\left(2,4,4\right);\left(4,4,2\right);\left(3,3,3\right)\)