cho a+b+c=a^2+b^2+c^2=3.tinh a^2018+b^2018+c^2018
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NH
0
ND
1
17 tháng 11 2018
\(a^3+b^3=c\left(3ab-c^2\right)\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2bc-2ca\right]=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)
Mà a + b + c = 3 nên a = b = c = 1
Khi đó \(A=672.\left(1+1+1\right)+2=672.3+2=2018\)
MT
0
LD
4 tháng 1 2019
\(P=\left(b^2c+abc\right)\left(a^2b+abc\right)\left(c^2a+abc\right)\)
\(=bc\left(a+b\right)\cdot ab\left(c+a\right)\cdot ca\left(b+c\right)\)
\(=\left(abc\right)^2\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Lại có:
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=0\)
\(\Leftrightarrow\left(a^2b+abc+a^2c\right)+\left(ab^2+b^2c+abc\right)+\left(bc^2+c^2a+abc\right)-abc=0\)
\(\Leftrightarrow a^2b+ca^2+ab^2+2abc+ac^2+b^2c+bc^2=0\)
\(\Leftrightarrow a^2\left(b+c\right)+a\left(b^2+2bc+c^2\right)+bc\left(b+c\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+a\left(b+c\right)^2+bc\left(b+c\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(a^2+ab+ca+bc\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)
\(\Rightarrow P=0\)
MH
0
NM
5 tháng 11 2018
a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)
VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)
Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)
