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\(a^2-2a+b^2+4b+4c^2-4c+6=0\)'
\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
b tự làm nốt nhé~
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)
\(M=x^3+3^3-x^3-54+x\)
\(M=x+27-54\)
\(M=x+27-54\)
\(M=7-27\)
\(M=-20\)
phân tích a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0
=>a=b=c(vì a+b+c khác 0)
thay a=b=c vào P
Nhận xét:\(\left(a+b\right)^3=a^3+b^3+3a^2b+3ab^2\)
=> \(a^3+b^3=\left(a+b\right)^3-3a^2b-3ab^2\)
ta có \(a^3+b^3+c^3-3abc\)
Thay vào biểu thức trên ta có:
\(\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
= \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
=\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
= \(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Vay \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Do \(a^3+b^3+c^3=3abc\)và theo đầu bài \(a+b+c\ne0\)nen \(a^2+b^2+c^2-ac-bc-ab=0\)
=> \(a=b=c\)
Vay N = \(\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]=0\)
Do \(a+b+c\ne0\) nên \(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-bc+c^2\right)+\left(c^2-ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\)
\(\Rightarrow\)\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
...
\(a)\) Ta có :
\(a+b+c=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^3=0^3\)
\(\Leftrightarrow\)\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(a+b+c=0\)\(\Rightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\)\(a^3+b^3+c^3+3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(a^3+b^3+c^3=3abc\) ( đpcm )
Vậy \(a^3+b^3+c^3=3abc\)
Chúc bạn học tốt ~
a, a+b+c=0 => a+b=-c
=>(a+b)3=(-c)3
=>a3+3a2b+3ab2+b3=-c3
=>a3+3ab(a+b)+b3=-c3
Mà a+b=-c
=>a3-3abc+b3=-c3
=>a3+b3+c3=3abc (đpcm)
b, \(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}\)
mà a3+b3+c3=3abc (bài a)
\(\Rightarrow P=\frac{3abc}{abc}=3\)
Vậy P=3
GT không hợp lí
Theo định lí cosi 3 số
a^3+b^3+c^3>=3*canbacba(a^3*b^3*c^3)
<=> a^3+b^3+c^3>=3abc
dấu"=" khi a=b=c
trái Gt a,b,c đôi một khác nhau
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Mà \(a+b+c\ne0\left(gt\right)\)
\(\Leftrightarrow a=b=c\)
Do đó:
\(A=\frac{a^2+2b^2+6c^2}{\left(a+b+c\right)^2}+2015=\frac{a^2+2a^2+6c^2}{\left(a+a+a\right)^2}+2015=\frac{9a^2}{9a^2}+2015=1+2015=2016\)