ý a bạn có chắc viết đề bài đúng không

đề bài đúng mà

a+b+c=0

=>a+b=-c; a+c=-b; b+c=-a

\(a^2-b^2-c^2\)

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

\(=a^2-\left\lbrack\left(b+c\right)^2-2bc\right\rbrack=a^2-\left(a^2-2bc\right)=2bc\)

\(b^2-a^2-c^2\)

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

\(=b^2-\left\lbrack\left(a+c\right)^2-2ac\right\rbrack=b^2-\left(b^2-2ac\right)=2ac\)

\(c^2-a^2-b^2\)

\(=c^2-\left\lbrack a^2+b^2\right\rbrack\)

\(=c^2-\left\lbrack\left(a+b\right)^2-2ab\right\rbrack=c^2-\left(c^2-2ab\right)=2ab\)

Ta có: \(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)

\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)

\(=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2bac}=\frac{\left(-c\right)^3-3ab\cdot\left(-c\right)+c^3}{2abc}=\frac{3abc}{2abc}=\frac32\)

a+b+c=0 suy ra a+b=-c ; a+c=-b ; b+c=-a 

bình phương hết lên ta có 

a^2+b^2+2ab=c^2 ; a^2+c^2+2ac=b^2 ; b^2+c^2+2bc=a^2

suy ra a^2+b^2-c^2=-2ab ; a^2+c^2-b^2=-2ac ; b^2+c^2-a^2=-2bc

thay vào B=-1/2(1/ab+1/bc+1/ac)=-1/2(c/abc+a/abc+b/abc)=0 do abc khác 0 và a+b+c=0

ko cần tích nhưng cần một lời cảm ơn

a³+a²c-abc+b²c+b³=a²(a+b+c)-a²b-abc+b²c+b³

=a².0+b²(a+b+c)-a²b-abc-b²a

=0+b².0-ab(a+b+c)=0+0-0=0

vậy a³+a²c-abc+b²c+b³=0

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

ab² hay là a²b²

Là a.b^2 nhé

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

Chứng minh: \(VP=\left(x+y\right)\left(x^2-xy+y^2\right)=x^3-x^2y+xy^2+x^2y-xy^2+y^3=x^3+y^3=VP\)

Áp dụng vào bài 

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Ta có \(a+b+c=0\Leftrightarrow-c=a+b\)

\(\Rightarrow c^2=\left(a+b\right)\left(a+b\right)=a^2+2ab+b^2\)

Xét \(a^3+b^3+a^2c+b^2c-abc\)

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

\(=a^3+b^3+c.c^2-3abc\)

\(=a^3+b^3+c^3-3abc\)

\(=a^3+a^2b+2a^2b+2ab^2+ab^2+b^3-3a^2b-3ab^2+c^3-3abc\)

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

\(=\left(a+b\right)\left(a^2+2ab+b^2\right)+c^3\) ( do a+b+c=0 )

\(=\left(a+b\right)\left[a\left(a+b\right)+b\left(a+b\right)\right]+c^3\)

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

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]=0\) ( do a+b+c=0 )

Vậy \(a^3+b^3+a^2c+b^2c-abc=0\)