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\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)[\left(a+b+c\right)^2-3ab-3ac-3bc]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right).2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right)[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)]\)
\(=\frac{1}{2}\left(a+b+c\right)[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2]\)
a)(x+a)(x+b)
=x(x+b)+a(x+b)
=x2+xb+ax+ab
=x2+(a+b).x+a.b
Vậy (x+a)(x+b)=x2+(a+b).x+a.b
b)(x+a)(x+b)(x+c)
=x(x+b)(x+c)+a(x+b)(x+c)
=(x2+xb)(x+c)+(ax+ab)(x+c)
=x2(x+c)+xb(x+c)+ax(x+c)+ab(x+c)
=x3+x2.c+x2.b+xbc+ax2+axc+abx+abc
=x3+(a+b+c).x2+(ab+bc+ca).x+abc
Vậy (x+a)(x+b)(x+c)=x3+(a+b+c).x2+(ab+bc+ca).x+abc
c)(a+b+c)(a2+b2+c2-ab-bc-ca)
=a(a2+b2+c2-ab-bc-ca)+b(a2+b2+c2-ab-bc-ca)+c(a2+b2+c2-ab-bc-ca)
=a3+ab2+ac2-a2.b-abc-a2.c+ba2+b3+bc2-ab2-b2.c-bca+ca2+cb2+c3-cab-bc2-c2.a
=a3+b3+c3 -abc-bca-cab
=a3+b3+c3 -3abc
Vậy (a+b+c)(a2+b2+c2-ab-bc-ca)=a3+b3+c3 -3abc
a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
a) Biến đổi vế trái ta có:
\(\left(x+a\right)\left(x+b\right)\)
= \(x^2+xb+xa+ab\)
= \(x^2+\left(a+b\right)x+ab=VP\)
Vậy đẳng thức đc CM
b) Biến đổi VT ta có:
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
= \(\left(x^2+xa+xb+ab\right)\left(x+c\right)\)
= \(x^3+x^2a+x^2b+x^2c+xab+xac+xbc+abc\)
= \(x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)= VP
Vậy đẳng thức đc CM
Câu 1:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
(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)\(\Rightarrowđpcm\)
(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) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)