1.(a+b+c)(a^2+b^2+c^2-ab-bc-ca)= a^3-b^3+c^3-3abc
2. (3a+2b-1)(a+5)-2b(a-2)=(3a+5)(a+3)+2(7b-10)
chứng minh các đẳng thức
1) a³ + b³ + c³ - 3abc
=(a + b)(a² - ab + b²) + c³ - 3abc
=(a + b)(a² - ab + b²) + c(a² - ab + b²) - 2abc - ca² - cb²
=(a + b + c)(a² - ab + b²) - (abc + b²c + bc² + ac² + abc + c²a) + c³ + ac² + bc²
=(a + b = c)(a² - ab + b²) - (a + b + c)(bc + ca) + c²(a + b + c)
=(a + b + c)(a² + b² + c² - ab - bc - ca)
2) \(\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)=\left(3a+5\right)\left(a-3\right)+2\left(7b-10\right)\left(1\right)\)
\(\Leftrightarrow3a^2+15a+2ab+10b-a-5-2ab+4b=3a^2+14a+15+14b-10\)
\(\Leftrightarrow3a^2+14a+14b-5=3a^2+14a+14b-5\)( đúng)
\(\Rightarrow\left(1\right)\) đúng (đpcm)
1) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\left(đpcm\right)\)
Cho ba cạnh của tam giác ABC là a,b,c Chứng minh tam giác ABC đều với các đẳng thức sau
a)(a+b+c)^2=3(ab+bc+ca)
b)a^3+b^3+c^3-3abc=0
c)(a+b)(b+c)(c+a)=8abc
chứng minh các hằng đẳng thức sau:
a,(a+b+c)(a2+b2+c2-ab-bc-ca)=a3+b3+c3-3abc
b,(3a+2b-1)(a+5)-2b(a-2)=(3a+5)(a+3)+2(7b-10)
chứng minh hằng đẳng thức
a)(a+b+c)^3 - a^3 -b^3 - c^3 = 3(a+b)(b+c)(c+a)
b) a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)
Giúp mình với nhé
Bài 3 chứng minh các đẳng thức sau:
a, a^3 + b^3 + c^3 - 3abc = ( a+b+c) ( a^2 + b^2 + c^2 -ab-bc-ca )
b, (a+b+c)^3 -a^3 -b^3 - c^3 = 3(a+b)(b+c)(c+a)
a^3+b^3+c^3-3abc=(a+b)^3-3a^2.b-3a.b^2-3abc=[(a+b)^3+c^3]-3ab(a+b+c)=(a+b+c).[(a+b)^2-c.(a+b)+c^2]-3ab(a+b+c)=(a+b+c).(a^2+2ab+b^2-ac-bc+c^2-3ab)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
Chứng minh các hằng đẳng thức:
a) (x+a)(x+b)=x2+(a+b)x+ab
b) (x+a)(x+b)(x+c)=x3+(a+b+c)x2+(ab+bc+ca)x+abc
a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)
B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
Chứng minh các hằng đẳng thức : a, \(\left(a+b+c\right)^3-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b, \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
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) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)
Chứng minh các đẳng thức sau :
a) ( a + b + c )^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)
b) a(b+c)^2 + b(a+c)^2 + c(a+b)^2 - 4abc = ( b+c)(c+a)(a+b)
Mấy bạn giải nhanh hộ mk a ! mk đang cần gấp
Chứng minh các đẳng thức sau:
(x+a)(x+b)(x+c)=x3+(a+b+c)x2+(ab+bc+ca)x+abc(a+b+c)(a2+b 2+c2-ab-bc-ca)=a3+b3+c3-3abc a2(b-c)+b2(c-a)+c2(a-b)=(a-b)(b-c)(a-c)
a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
a,\(\left(x+a\right)\left(x+b\right)\left(x+c\right)=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
\(VT=\left[\left(x+a\right)\left(x+b\right)\right]\left(x+c\right)\)
\(=\left(x^2+bx+ax+ab\right)\left(x+c\right)\)
\(=\left[x^2+x\left(a+b\right)+ab\right]\left(x+c\right)\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\left(dpcm\right)\)
b,\(VP=a^3+b^3+c^3-3abc\)
\(=a^3+b^3+c^3+3a^2b-3a^2b+3ab^2-3ab^2-3abc\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)-3a^2b-3ab^2+c^3+3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
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