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Bài 1:
a ) a.( b2 + c2 ) + b.( a2 + c2 ) + c.( a2 + b2 ) + 2abc
= ab2 + ac2 + a2b + bc2 + a2c + b2c + 2abc
= ( ab2 + a2b ) + ( ac2 + bc2 ) + ( a2c + 2abc + b2c )
= ab.( a + b ) + c2.( a + b ) + c.( a2 + 2ab + b2 )
= ab.( a + b ) + c2.( a + b )v + c.( a + b)2
= ( a + b ).[ ( ab + c2 + c. ( a + b ) ]
= ( a + b ).( ab + c2 + ac + bc )
= ( a + b ).[ ( ab + ac ) + ( c2 + bc) ]
= ( a + b ).[ a.( b + c ) + c.( b + c ) ]
= ( a + b ).( b + c ).( a + c )
b) ab.( a + b ) - bc.( b + c ) + ac.( a - c )
= ab.( a + b ) - bc.( b + c ) + ac.[ ( a + b ) - ( b + c ) ]
= ab.( a + b ) - bc. ( b + c ) + ac.( a + b ) - ac.( b + c )
= ab.( a + b ) + ac.( a + b ) - bc.( b + c ) - ac.( b + c )
= ( a + b ).( ab + ac ) + ( b + c ).( -bc - ac )
= ( a + b ).a.( b + c ) - ( b + c ).c.( a + b )
= ( a + b ).( b + c ).( a - c )
c) ( x2 + x )2 + 2.( x2 + x ) - 3
Đặt x2 + x = a
Khi đó đa thức trở thành:
a2 + 2a - 3
= a2 + 3a - a - 3
= a.( a + 3 ) - ( a + 3 )
= ( a - 1 ).( a - 3 )
\(\Rightarrow\) ( x2 + x - 1 ).( x2 + x - 3 )
B2
ab.( a - b ) + bc.( b - c ) + ca.( c - a ) = 0
\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.[ ( a - b ) + ( b - c ) ] = 0
\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.( a - b ) - ca.( b - c ) = 0
\(\Leftrightarrow\)ab.( a - b ) - ca.( a - b ) + bc.( b - c ) - ca.( b - c ) = 0
\(\Leftrightarrow\) ( a - b ).( ab - ca ) + ( b - c ).( bc - ca ) = 0
\(\Leftrightarrow\) ( a - b ).a.( b - c ) - ( b - c ).c.( a - b ) = 0
\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0
\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0
\(\Leftrightarrow\) a = b , b = c , a = c
\(\Rightarrow\) a = b = c
Ta có: a3+b3+c3=3abc
<=> (a+b+c)(a2+b2+c2-ab-bc-ca)=0
<=> (a+b+c)(2a2+2b2+2c2-2ab-2bc-2ca)=0
<=> (a+b+c)[(a-b)2+(b-c)2+(c-a)2 ] = 0
<=> \(\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Vì a,b,c phân biệt nên a+b+c=0 => \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(c+a\right)\\c=-\left(a+b\right)\end{cases}}\)(*)
Lại có: \(M=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}\)
Thay (*) vào M ta được:
\(M=\frac{-\left(b+c\right)b^2}{\left(b+c\right)^2+\left(b+c\right)\left(b-c\right)}+\frac{-\left(c+a\right)c^2}{\left(c+a\right)^2+\left(c+a\right)\left(c-a\right)}+\frac{-\left(a+b\right)a^2}{\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)}\)
\(=\frac{-\left(b+c\right)b^2}{\left(b+c\right)\left(b+c+b-c\right)}+\frac{-\left(c+a\right)c^2}{\left(c+a\right)\left(c+a+c-a\right)}+\frac{-\left(a+b\right)a^2}{\left(a+b\right)\left(a+b+a-b\right)}\)
\(=\frac{-\left(b+c\right)b^2}{2b\left(b+c\right)}+\frac{-\left(c+a\right)c^2}{2c\left(c+a\right)}+\frac{-\left(a+b\right)a^2}{2a\left(a+b\right)}\)
\(=\frac{-b}{2}-\frac{c}{2}-\frac{a}{2}=\frac{-\left(b+c+a\right)}{2}\)
Mà a+b+c=0
=> M=0
Vậy M=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)\)
\(-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\Rightarrow\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)
\(\left(a-b\right)^2>=0\Rightarrow a^2-2ab+b^2>=0\Rightarrow a^2+b^2>=2ab\)
tương tự \(a^2+c^2>=2ac;b^2+c^2>=2bc\)
\(\Rightarrow a^2+b^2+a^2+c^2+b^2+c^2>=2ab+2ac+2bc\Rightarrow2\left(a^2+b^2+c^2\right)>=2\left(ab+ac+bc\right)\)
\(\Rightarrow a^2+b^2+c^2.=ab+ac+bc\)dấu = xảy ra khi a=b=c
mà nếu \(a^2+b^2+c^2-ab-ac-bc=0\Rightarrow a^2+b^2+c^2=ab+ac+bc\Rightarrow a=b=c\)
th1:a+b+c=0
\(\Rightarrow a+b=-c;a+c=-b;b+c=-a\)
\(M=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}=\frac{ab^2}{a^2+b^2-\left(-c\right)^2}+\frac{bc^2}{b^2+c^2-\left(-a\right)^2}+\frac{ca^2}{c^2+a^2-\left(-b\right)^2}\)
\(=\frac{ab^2}{a^2+b^2-\left(a+b\right)^2}+\frac{bc^2}{b^2+c^2-\left(b+c\right)^2}+\frac{ca^2}{c^2+a^2-\left(c+a\right)^2}\)
\(=\frac{ab^2}{a^2+b^2-a^2-2ab-b^2}+\frac{bc^2}{b^2+c^2-b^2-2bc-c^2}+\frac{ca^2}{c^2+a^2-c^2-2ac-a^2}\)
\(=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ac}=\frac{b}{-2}+\frac{c}{-2}+\frac{a}{-2}=\frac{a+b+c}{-2}=\frac{0}{-2}=0\)
th2:a=b=c tự lm nhá
2, (trích đề thi học sinh giỏi Bến Tre-1993)
\(a^3+a^2b+ca^2+b^3+ab^2+b^2c+c^3+c^2b+c^2a=a^2\left(a+b+c\right)+b^2\left(a+b+c\right)+c^2\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
mà a+b+c=0 => (a+b+c)(a2+b2+c2)=0
=> đpcm
*bài này tui làm tắt, không hiểu ib
Vừa lm xog bị troll chứ, tuk quá
\(x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\)
\(\Leftrightarrow\frac{x\left(b^2-x^2\right)\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}-\frac{a^2x\left(b^2-x^2\right)\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}-\frac{b^2\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}+\frac{a\left(b^2-x^2\right)\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}=\frac{x^2\left(b^2-x^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}\)
Khử mẫu :
\(\Leftrightarrow2x^3b^2-xb^4-x^5-2a^2x^3b^2+a^2xb^4+a^2x^5-b^2x^2+b^4+2ab^2x^2-ab^4-ax^4=x^2b^2-x^4\)
Tự xử nốt, lm bài này muốn phát điên mất.
Biến đổi tương đương:
\(\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3\left(ab+ac+bc\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b=c\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{ab+ac+bc}\ge3\)
b/ \(VT=\frac{\left(a+b+c\right)^2}{ab+ac+bc}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}=\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}\)
\(\Rightarrow VT\ge\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+2\sqrt{\frac{\left(a+b+c\right)^2\left(ab+ac+bc\right)}{9\left(ab+ac+bc\right)\left(a+b+c\right)^2}}\ge\frac{8.3}{9}+\frac{2}{3}=\frac{10}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
