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\(a^3+b^3+c^3-3abc\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-\left(3a^2b+3ab^2+3abc\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)\(\left(đpcm\right)\)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Sửa đề : CM \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Ta có : \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3a^2b-3b^2a-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VP\)
\(\left(đpcm\right)\)
* Đặt tên các biểu thức theo thứ tự là A,B,C,D,E.
Câu a)
Theo hằng đẳng thức đáng nhớ ta có:
\(a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)\)
\(=(a+b+c)^3-3[ab(a+b)+bc(b+c)+ca(c+a)+2abc]\)
\(=(a+b+c)^3-3[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)-abc]\)
\(=(a+b+c)^3-3[(a+b+c)(ab+bc+ac)]+3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)^3-3(ab+bc+ac)(a+b+c)\)
\(=(a+b+c)[(a+b+c)^2-3(ab+bc+ac)]\)
\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\) (*)
Do đó:
\(A=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)
Câu b)
\(x^3-y^3+z^3+3xyz=x^3+(-y)^3+z^3-3x(-y)z\)
Sử dụng kết quả (*) của câu a. Với \(a=x, b=-y, c=z\)
\(\Rightarrow x^3+(-y)^3+z^3-3x(-y)z=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)
Mặt khác xét mẫu số:
\((x+y)^2+(y+z)^2+(x-z)^2=x^2+2xy+y^2+y^2+2yz+z^2+x^2-2xz+z^2\)
\(=2(x^2+y^2+z^2+xy+yz-xz)\)
Do đó: \(B=\frac{(x-y+z)(x^2+y^2+z^2+xy+yz-xz)}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{x-y+z}{2}\)
Câu c) Sử dụng kết quả (*) của phần a:
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\)
Và mẫu số:
\((x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2-xy-yz-xz)\)
Do đó: \(C=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{2(x^2+y^2+z^2-xy-yz-xz)}=\frac{x+y+z}{2}\)
Câu d)
Xét tử số:
\(a^2(b-c)+b^2(c-a)+c^2(a-b)\)
\(=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)\)
\(=(b-c)(a^2-b^2)-(b^2-c^2)(a-b)\)
\(=(b-c)(a-b)(a+b)-(b-c)(b+c)(a-b)\)
\(=(a-b)(b-c)[a+b-(b+c)]=(a-b)(b-c)(a-c)\) (1)
Xét mẫu số:
\(a^4(b^2-c^2)+b^4(c^2-a^2)+c^4(a^2-b^2)\)
\(=a^4(b^2-c^2)-b^4[(b^2-c^2)+(a^2-b^2)]+c^4(a^2-b^2)\)
\(=(a^4-b^4)(b^2-c^2)-(b^4-c^4)(a^2-b^2)\)
\(=(a^2-b^2)(a^2+b^2)(b^2-c^2)-(b^2-c^2)(b^2+c^2)(a^2-b^2)\)
\(=(a^2-b^2)(b^2-c^2)[a^2+b^2-(b^2+c^2)]\)
\(=(a^2-b^2)(b^2-c^2)(a^2-c^2)\)
\(=(a-b)(b-c)(a-c)(a+b)(b+c)(c+a)\)(2)
Từ (1)(2) suy ra \(D=\frac{1}{(a+b)(b+c)(c+a)}\)
Câu e)
Theo phần d ta có:
\(TS=(a-b)(b-c)(a-c)\)
\(MS=ab^2-ac^2-b^3+bc^2\)
\(=b^2(a-b)-c^2(a-b)=(a-b)(b^2-c^2)=(a-b)(b-c)(b+c)\)
Do đó: \(E=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}=\frac{a-c}{b+c}\)
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) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)
a. \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)
b. \(\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x-y\right)^2-\left(x-y\right)z+z^2\right]+3xy\left(x-y+z\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{2\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x+y-z\right)\left(2x^2+2y^2+2z^2+2xy+2yz-2zx\right)}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}=\dfrac{x-y+z}{2}\)
a) Biến đổi vế phải ta có:
\(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)=a^3+b^3=VT\)
Vậy đẳng thức trên đc chứng minh
b) Sai đề sửa lại
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Biến đổi vế trái ta có:
\(a^3+b^3+c^3-3abc\)
\(=\left(a^3+b^3\right)+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=VP\)
Vậy đẳng thức trên đc chứng minh
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
Giải:
Ta có: \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(=VP\) (Đpcm)
Ta có:
\(VP=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc+a^2c+b^2c+c^3-abc-bc^2-ac^2\)
\(=a^3+b^3+c^3-3abc=VT\)
\(\rightarrow\) đpcm
Chúc bạn học tốt!!!