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

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

=>a=-b hoặc a=-c hoặc b=-c (1)

=>a=1 hoăc b=1 hoặc c=1 (2)

từ 1 và 2 => Q=1

giúp mình với mình cần gấp

gt \(\Rightarrow\left\{{}\begin{matrix}b\left(a^2+2ac+c^2\right)+ac\left(a+c\right)+b^2\left(a+c\right)=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\\a^{2013}+b^{2013}+c^{2013}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=0\Rightarrow a^{2013}+b^{2013}=0\\b+c=0\Rightarrow b^{2013}+c^{2013}=0\\a+c=0\Rightarrow a^{2013}+c^{2013}=0\end{matrix}\right.\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow Q=1\)

cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)

\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)

\(=x-1=2013-1=2012\)

nhiều quá

lớp 7 hả

cái chỗ a+c+1 la "ac+c+1" nha, mình viết nhầm

ta có: \(\frac{2013a^2bc}{ab+2013a+2013}\)= \(\frac{2013.ab.ac}{ab+ab.ac+abc}\)= \(\frac{2013.ab.ac}{ab.\left(ac+c+1\right)}\)= \(\frac{2013ac}{ac+c+1}\)

\(\frac{ab^2c}{bc+b+2013}\)= \(\frac{abc.b}{bc+b+abc}\)= \(\frac{2013b}{b\left(ac+c+1\right)}\)= \(\frac{2013}{ac+c+1}\)

\(\frac{abc^2}{ac+c+1}\)= \(\frac{abc.c}{ac+c+1}\)= \(\frac{2013c}{ac+c+1}\)

Cộng cả 3 phân thức cùng mẫu thức ta có phân thức cuối cùng là:

P=\(\frac{2013.\left(ac+c+1\right)}{ac+c+1}\)=2013

1b/

Áp dụng BĐT Cô-si :
\(\sqrt{\frac{b+c}{a}}\le\frac{\frac{b+c}{a}+1}{2}=\frac{\frac{a+b+c}{a}}{2}=\frac{a+b+c}{2a}\)

\(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)

Chứng minh tương tự:

\(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\); \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)

Cộng theo vế ta được :

\(VT\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Dấu "=" không xảy ra nên \(VT>2\).

2a/ Chắc là tính GT của \(x+y\).

\(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\)

\(\Leftrightarrow\left(x-\sqrt{x^2+2013}\right)\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2013}=\sqrt{x^2+2013}-x\)

Do vai trò \(x,y\) là như nhau nên thiết lập tương tự ta có :

\(x+\sqrt{x^2+2013}=\sqrt{y^2+2013}-y\)

Cộng theo vế 2 pt ta được :

\(x+y+\sqrt{x^2+2013}+\sqrt{y^2+2013}=\sqrt{x^2+2013}+\sqrt{y^2+2013}-x-y\)

\(\Leftrightarrow2\left(x+y\right)=0\)

\(\Leftrightarrow x+y=0\)

Vậy....

2b/

Đặt \(A=5a^2+15ab-b^2\) và \(B=3a+b\)

Ta có \(B^2=\left(3a+b\right)^2=9a^2+6ab+b^2\)

Lấy \(A+B^2=5a^2+15a-b^2+9a^2+6ab+b^2\)

\(A+B^2=14a^2+21ab\)

\(A+B^2=7\left(2a+3ab\right)⋮7\)

Mà \(A⋮7\) ( vì \(A⋮49\) ) nên \(B^2⋮7\)

Vì 7 nguyên tố nên \(B⋮7\) ( đpcm )

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(\frac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right)=0\)

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

\(\Leftrightarrow\left(a^{2013}+b^{2013}\right)\left(b^{2013}+c^{2013}\right)\left(c^{2013}+a^{2013}\right)=0\)

\(\Rightarrow P=\frac{17}{25}\)