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Giả sử a^3+b^3+c^3=3abc

<=> a^3+b^3+c^3-3abc=0

<=> (a+b)^3 -3ab(a+b) -3abc +c^3=0

<=>[(a+b)^3+c^3] -3ab(a+b+c) =0

<=> (a+b+c)[(a+b)^2-(a+b)c+c^3] -3ab(a+b+c)=0

<=> (a+b+c)[(a+b)^2-(a+b)c+c^3-3ab]=0 

vì a+b+c =0 => đpcm

a) sau khi nhân vô + rút gọn ( câu này gg có á)

P = a³ + b³ + c³ - 3abc

b) a³ + b³ + c³ = 3abc?

a³ + b³ + c³ - 3abc = 0

theo câu b)

(a + b + c)(a² + b² + c² - ab - bc - ca) =0

\(\Rightarrow\) a+b+c=0 hoặc

a² + b² + c² - ab - bc -ca = 0

a² - 2ab +b² +b² - 2bc + c² + c² - 2ac +a² =0

(a-b)² + (b-c)² + (c-a)² = 0

\(\Rightarrow\) a=b=c

Sửa đề: Cho \(x+y=a;x^2+y^2=b;x^3+y^3=c\)

Chứng minh: \(a^3-2ab+2c=0\)

Giải:

Ta có:

\(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)

\(=x^3+y^3+3xy\left(x+y\right)-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)

\(=3\left(x^3+y^3\right)+3\left(x+y\right)\left(xy-x^2-y^2\right)=3\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(x+y\right)\left(xy-x^2-y^2\right)\)

\(=3\left(x+y\right)\left(x^2-xy+y^2+xy-x^2-y^2\right)=3\left(x+y\right).0\)

\(=0\) (đpcm)

Ta có:\(x+y=a\)

=>\(x^2+2xy+y^2=a^2\)

=>\(x^2+y^2=a^2-2xy=a^2-2b\left(đpcm\right)\)

Ta lại có:\(x^3+3x^2y+3xy^2+y^3=a^3\)

=>\(x^3+y^3+3xy\left(x+y\right)=a^3\)

=>\(x^3+y^3=a^3-3xy\left(x+y\right)=a^3-3ab\left(đpcm\right)\)

b)\(a+b+c=0\) =>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\) =>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\) =>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\) =>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)

\(a+b+c=0\)

=>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\)

=>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

=>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\)

=>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\left(1\right)\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\left(2\right)\)

\(\left(1\right)\left(2\right)\RightarrowĐpcm\)

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