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Biến đổi vế trái:
a + b + c 3 = a + + c 3 = a + b 3 +3 a + b 2 c+3(a+b) c 2 + c 3
= a 3 + 3 a 2 b + 3a b 2 + b 3 + 3( a 2 + 2ab + b 2 )c + 3a c 2 + 3b c 2 + c 3
= a 3 + 3 a 2 b + 3a b 2 + b 3 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2 + 3b c 2 + c3
= a 3 + b 3 + c 3 + 3 a 2 b + 3a b 2 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2 + 3b c 2
= a 3 + b 3 + c 3 + (3 a 2 b + 3a b 2 ) +( 3 a 2 c + 3abc)+ (3abc + 3 b 2 c)+(3a c 2 + 3b c 2 )
= a 3 + b 3 + c 3 + 3ab(a + b) + 3ac(a + b) + 3bc(a + b) + 3 c 2 (a + b)
= a 3 + b 3 + c 3 + 3(a + b)(ab + ac + bc + c 2 )
= a 3 + b 3 + c 3 + 3(a + b)[a(b + c) + c(b + c)]
= a 3 + b 3 + c 3 + 3(a + b)(b + c)(a + c) (đpcm)
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#)Giải :
Ta có : \(\left(a+b+c\right)^3\)
\(=\left(\left(a+b\right)+c\right)^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+3\left(a+b\right)\left(ab+c\left(a+b+c\right)\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Hay chính là \(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Rightarrowđpcm\)
ta có:
VT=(a+b+c)^3=[(a+b)+c]^3
=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3ab(a+b)+3c(a+b+c)(a+b)
=a^3+b^3+c^3+3(a+b)(ab+ac+cb+c^2)
=a^3+b^3+c^3+3(a+b)(b+c)(c+a)
=>VT=VP( đpcm)
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\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)
b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24
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a) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2+bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)
b) Ta có:
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi
a) Ta có: (a + b + c)3 - a3 - b3 - c3 = [ (a + b + c)3 - a3 ] - ( b3 + c3)
= (a + b + c - a) ( a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2) - (b + c) ( b2 - bc + c3)
= (b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac) - (b + c) ( b2 - bc + c3)
= ( b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac - b2 + bc - c3)
= ( b + c) ( 3a2 + 3ab + 3bc + 3ac)
= 3 (b + c) [a (a + b) + c (a + b)]
= 3 (b + c) (a + b) (a + c) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$a^3+b^3=2(c^3-8d^3)$
$a^3+b^3+c^3+d^3=c^3+d^3+2(c^3-8d^3)$
$=3c^3-15d^3=3(c^3-5d^3)\vdots 3$
Khi đó:
$(a+b+c+d)^3=(a+b)^3+(c+d)^3+3(a+b)(c+d)(a+b+c+d)$
$=a^3+b^3+c^3+d^3+3ab(a+b)+3cd(c+d)+3(a+b)(c+d)(a+b+c+d)\vdots 3$ do:
$a^3+b^3+c^3+d^3\vdots 3$
$3ab(a+b)\vdots 3$
$3cd(c+d)\vdots 3$
$3(a+b)(c+d)(a+b+c+d)\vdots 3$
Vậy:
$(a+b+c+d)^3\vdots 3$
$\Rightarrow a+b+c+d\vdots 3$
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\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
$a^3+b^3+c^3=3abc$
$\Leftrightarrow (a+b)^3-3ab(a+b)+c^3-3abc=0$
$\Leftrightarrow [(a+b)^3+c^3]-[3ab(a+b)+3abc]=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2-3ab]=0$
$\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$
$\Rightarrow a+b+c=0$ hoặc $a^2+b^2+c^2-ab-bc-ac=0$
Xét TH $a^2+b^2+c^2-ab-bc-ac=0$
$\Leftrightarrow 2(a^2+b^2+c^2)-2(ab+bc+ac)=0$
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow a-b=b-c=c-a=0$
$\Leftrightarrow a=b=c$
Vậy $a^3+b^3+c^3=3abc$ khi $a+b+c=0$ hoặc $a=b=c$
Áp dụng vào bài:
Nếu $a+b+c=0$
$A=\frac{-c}{c}+\frac{-b}{b}+\frac{-a}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$
$P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2+2+2=6$