Question

[quangduy]

Tính góc A của tam giác ABC biết:

a) \(\dfrac{b^3+c^3-a^3}{b+c-a}=a^2\)

b) \(cosB=\dfrac{\left(a+b\right)\left(b+c-a\right)\left(c+a-b\right)}{2abc}\)

c) \(a^4-2\left(b^2+c^2\right)a^2+b^4+b^2c^2+c^4=0\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\Leftrightarrow\dfrac{\left(b+c\right)^3-a^3-3bc\left(b+c\right)}{b+c-a}=a^2\)

\(\Leftrightarrow a^2=\left(b+c\right)^2+a\left(b+c\right)+a^2-\dfrac{3bc\left(b+c\right)}{b+c-a}\)

\(\Leftrightarrow\left(b+c\right)^2+a\left(b+c\right)-\dfrac{3bc\left(b+c\right)}{b+c-a}=0\)

\(\Leftrightarrow\left(b+c\right)\left(b+c-\dfrac{3bc}{b+c-a}+a\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left(b+c-a\right)-3bc+a\left(b+c-a\right)=0\)

\(\Leftrightarrow b^2+2bc+c^2-ab-ac-3bc+ab+ac-a^2=0\)

\(\Leftrightarrow b^2+c^2-a^2-bc=0\)

\(\Leftrightarrow a^2=b^2+c^2-bc\)

\(cosA=\dfrac{b^2+c^2-a^2}{2\cdot b\cdot c}=\dfrac{1}{2}\)

nên góc A=30 độ

b: \(cosB=\dfrac{\left(a+b\right)\left(b+c-a\right)\left(c+a-b\right)}{2bac}\)

=>\(\dfrac{\left(a+b\right)\left[c-\left(a-b\right)\right]\left[c+\left(a-b\right)\right]}{2abc}=\dfrac{a^2+c^2-b^2}{2ac}\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\cdot\left[c^2-\left(a-b\right)^2\right]}{b}=a^2+c^2-b^2\)

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

\(\Leftrightarrow ac^2+bc^2-\left(a^2-b^2\right)\left(a-b\right)=a^2b+c^2b-b^3\)

\(\Leftrightarrow ac^2+bc^2-a^3+a^2b+ab^2-b^3=a^2b+c^2b-b^3\)

\(\Leftrightarrow ac^2+bc^2-a^3+ab^2=c^2b\)

\(\Leftrightarrow ac^2+bc^2-a^3-ab^2-c^2b=0\)

\(\Leftrightarrow c^2\left(a+b\right)-a\left(a^2+b^2\right)-c^2b=0\)

=>c^2*a-a(a^2+b^2)=0

=>a(c^2-a^2-b^2)=0

=>c^2=a^2+b^2

=>góc A=90 độ