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b) \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Biến đổi VT ta có :
+) \(a^3+b^3+c^3=ab+bc+ca\)
\(\Leftrightarrow3a^3+3b^3+3c^3=3ab+3bc+3ca\)
\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=0\)
\(\Rightarrow a=b=c\)
< => VT = VP
=> đpcm
\(VP=\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3=VT\)
(a+b+c)3 = (a + b)3 + c3 + 3(a+b)c.(a+b+c)
= a3 + b3 + 3ab.(a+b) + c3 + 3(a+b)c(a+b+c) = a3 + b3 + c3 + 3(a+b). (ab + ac + bc + c2 )
= a3 + b3 + c3 + 3.(a+b). [a(b+c) + c.(b+c)] = a3 + b3 + c3 + 3(a+b).(a+c).(b+c)\(\Rightarrowđpcm\)
(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)2c+3(a+b)c2=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)=a3+b3+c3+3(a+b)(c(a+b+c)+ab)=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)2c+3(a+b)c2=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)=a3+b3+c3+3(a+b)(c(a+b+c)+ab)=a3+b3+c3+3(a+b)(b+c)(c+a)(Cái trong ngoặc bạn tự phân tích đa thức thành nhân tử 1 cách dễ dàng.
Cách2:Xét hiệu :
\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=\left(a^3+3a^2b+3b^2a+b^3\right)+3c\left(a^2+2ab+b^2\right)+3c^2\left(a+b\right)+c^3\)
\(=a^3+3a^2b+3b^2a+b^3+3a^2c+6abc+3b^2c+3ac^2+3bc^2+c^3\)
\(=a^3+b^3+c^3+\left(3a^2b+3b^2a+3b^2c+3c^2b+3a^2c+3c^2a+6abc\right)\)
\(=a^3+b^3+c^3+3\left(a^2b+b^2a+b^2c+c^2b+a^2c+c^2a+2abc\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(VP=a^3+b^3+c^3+\left(3ab+3ac+3b^2+3bc\right)\left(c+a\right)\)a)
= a3 + b3 + c3 + 3abc + 3ac2 + 3b2c + 3bc2 + 3a2b + 3a2c + 3b2a + 3abc
= ( a + b )3 + 3( a+b)2c + 3(a+b)c2 + c3
= (a+b+c)3
\(\left(a+b+c\right)^3\)
\(=\left[\left(a+b\right)+c\right]^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[ab+\left(a+b\right)c+c^2\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=VP\left(đpcm\right)\)
a) Biến đổi VT . Mẫu chung là ( a + 2b )( a - 2b )
\(VT=\frac{a+2b-6b-2\left(a-2b\right)}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 1 )
Biến đổi VP
\(-\frac{1}{2a}\left(\frac{a^2+4b^2}{a^2-4b^2}+1\right)=-\frac{1}{2a}\cdot\frac{a^2+4b^2+a^2-4b^2}{a^2-4b^2}\)
\(=-\frac{1}{2a}\cdot\frac{2a^2}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 2 )
Từ ( 1 ) và ( 2 ) => VT = VP ( đpcm )
b) \(a^3+b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)^3\)
<=> \(b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)^3=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)-a^3\)( * )
Biến đổi VT của ( * ) ta có :
\(VT=\left[b+\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right]\left[b^2-\frac{b^2\left(2a^3+b^3\right)}{a^3-b^3}+\frac{b^2\left(2a^3+b^3\right)^2}{\left(a^3-b^3\right)^2}\right]\)
\(=\frac{3a^3b}{a^3-b^3}\cdot\frac{3a^6b^2+3a^3b^5+3b^8}{\left(a^3-b^3\right)^2}\)
\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 1 )
\(VP=\left[\frac{a\left(a^3+2b^3\right)}{a^3-b^3}-a\right]\left[\frac{a^2\left(a^3+2b^3\right)^2}{\left(a^3-b^3\right)^2}+\frac{a^2\left(a^3+2b^3\right)}{a^3-b^3}+a^2\right]\)
\(=\frac{3ab^3}{a^3-b^3}\cdot\frac{3a^8+3a^5b^3+3a^2b^6}{\left(a^3-b^3\right)^2}\)
\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 2 )
Từ ( 1 ) và ( 2 ) => VT = VP => ( * ) đúng
=> Hằng đẳng thức đúng
(a+b+c)3=a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3c2a+3ca2+6abc
a3+b3+c3+3(a+b)(b+c)(c+a)=a3+b3+c3+3(a+b)(ab+ac+bc+c2)=a3+b3+c3+3a2b+3a2c+3abc+3ac2+3ab2+3abc+3b2c+3bc2=a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3c2a+3ca2+6abc
=>(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
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