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\(VP=a^3+b^3+c^3+\left(3ab+3ac+3b^2+3bc\right)\left(c+a\right)\)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+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 :
a) \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)\left(ac+bc+c^2\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
b) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)
Câu a : Ta có : \(x^3+x^2z+y^2z-xyz+y^3=0\)
\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z+y^2z-xyz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)
\(\Leftrightarrow x+y+z=0\) ( đpcm )
Câu b : \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Câu c : Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a+b+c=0\) ( đúng )
(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)c(a+b+c)(a+b+c)3=((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+3ab(a+b)+c3+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)(ab+c(a+b+c))=a3+b3+c3+3(a+b)(ab+c(a+b+c))
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)(a+c)(b+c)=a3+b3+c3+3(a+b)(a+c)(b+c)
~~
(a+b+c)3=(a+b)3+3(a+b)2c+3(a+b)c2+c3
=a3+b3+c3+3a2b+3ab2+3(a2+2ab+b2)c+3ac2+3bc2
=a3+b3+c3+3a2b+3ab2+3a2c+6abc+3b2c+3ac2+3bc2
(a+b)(b+c)(a+c)=(a+b)(b+c)a+(a+b)(b+c)c
=(a+b)ab+(a+b)ac+(a+b)bc+(a+b)c2
=a2b+ab2+a2c+abc+abc+b2c+ac2+bc2
=>3(a+b)(b+c)(c+a)=3a2b+3ab2+6abc+3b2c+3ac2+3bc2
=>(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
=>đpcm
1. biến đổi vế trái
= a2x2 + a2y2 + b2x2 + b2y2
= (ax -by)2 + (bx+ ay)2 - 2abxy + 2abxy
= (ax -by)2 + ( bx + ay)2 = vế phải( dpcm)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)
a)
Ta có
a chia 5 dư 4
=> a=5k+4 ( k là số tự nhiên )
\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)
Vì 25k^2 chia hết cho 5
40k chia hết cho 5
16 chia 5 dư 1
=> đpcm
2) Ta có
\(12=\frac{5^2-1}{2}\)
Thay vào biểu thức ta có
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)
\(\Rightarrow P=\frac{5^{16}-1}{2}\)
3)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
<=>(a+b+c)3-a3-b3-c3=3(a+b)(b+c)(c+a) (1)
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+2ca+a2+ab+ac+a2)-(b+c)(b2-bc+c2)
=(b+c)(3a2+b2+c2+3ab+3ac+2bc)-(b+c)(b2-bc+c2)
=(b+c)(3a2+b2+c2+3ab+3ac+2bc-b2+bc-c2)
=(b+c)(3a2+3ab+3ac+3bc)
=3(b+c)](a2+ab)+(ac+bc)]
=3(b+c)[a(a+b)+c(a+b)]
=3(b+c)(a+c)(a+b)
=>(1) đúng => đpcm
biến đổi vế trái :
\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3\left(a^2+2ab+b^2\right)c+3ac^2+3bc^2+c^3\)
\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3ac\left(a+b\right)+3bc\left(a+b\right)+3c^2\left(a+b\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Vế trái bằng vế phải đẳng thức được chứng minh.
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)