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\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
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. \(\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x-y\right)^2-\left(x-y\right)z+z^2\right]+3xy\left(x-y+z\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{2\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x+y-z\right)\left(2x^2+2y^2+2z^2+2xy+2yz-2zx\right)}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}=\dfrac{x-y+z}{2}\)
* 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: \(E=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}=\frac{a-c}{b+c}\)
\(A=\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\dfrac{\left(a^2-b^2\right)\left(b+c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(b^2-c^2\right)}\)
\(\dfrac{\left(a-b\right)\left(a+b\right)\left(b+c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}\)
\(=\dfrac{a-c}{b+c}\)
Vậy..
Ta có
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}=\frac{a^2+ab-bc-ab}{\left(a+b\right)\left(a+c\right)}=\frac{a\cdot\left(a+b\right)-b\cdot\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}=\frac{a}{a+c}-\frac{b}{a+b}\left(1\right)\)
tương tự
\(\frac{b^2-bc}{\left(a+b\right)\left(b+c\right)}=\frac{b}{a+b}-\frac{c}{b+c}\left(2\right)\)
\(\frac{c^2-ab}{\left(c+a\right)\left(b+c\right)}=\frac{c}{c+b}-\frac{a}{a+b}\left(3\right)\)
Cộng (1);(2) và (3) ta có
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}=\frac{a}{a+c}-\frac{b}{a+b}+\frac{b}{a+b}-\frac{c}{b+c}+\frac{c}{c+b}-\frac{a}{a+b}=0 \)
Lời giải:
\(a^2(b-c)+b^2(c-a)+c^2(a-b)=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)\)
\(=(a^2-b^2)(b-c)-(b^2-c^2)(a-b)\)
\(=(a-b)(a+b)(b-c)-(b-c)(b+c)(a-b)\)
\(=(a-b)(b-c)(a+b-b-c)=(a-b)(b-c)(a-c)\)
Và:
\(ab^2-ac^2-b^3+bc^2=(ab^2-b^3)-(ac^2-bc^2)\)
\(=b^2(a-b)-c^2(a-b)=(b^2-c^2)(a-b)=(b-c)(b+c)(a-b)\)
Do đó: \(P=\frac{(a-b)(b-c)(a-c)}{(b-c)(b+c)(a-b)}=\frac{a-c}{b+c}\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(a-b\right)-b^2\left(b-c\right)+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(ab^2-ac^2-b^3+bc^2\)
\(=b^2\left(a-b\right)-c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
Vậy \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{a-c}{b+c}\)
Có a2(b-c) + b2(c-a) + c2(a-b)
= a2(b-c) - b2(a-c) + c2(a-b)
= a2(b-c) - b2(b-c+a-b) + c2(a-b)
= a2(b-c) - b2(b-c) - b2(a-b) + c2(a-b)
=[a2(b-c) - b2(b-c)] - [b2(a-b) - c2(a-b)]
=(b-c)(a2-b2) - (a-b)(b2-c2)
=(b-c)(a-b)(a+b) - (a-b)(b-c)(b+c)
=(b-c)(a-b)[(a+b)-(b+c)]
=(b-c)(a-b)(a-c)
Có ab2 - ac2 - b3 + bc2
= (ab2-ac2) - (b3-bc2)
=a(b2-c2) - b(b2-c2)
=(b2-c2)(a-b)
=(b-c)(b+c)(a-b)
Có a2(b-c) + b2(c-a) + c2(a-b) / ab2 - ac2 - b3 + bc2
= (b-c)(a-b)(a-c) / (b-c)(b+c)(a-b)
= (a-c) / (b+c)
AD phân tích đa thức thành nhân tử ở tử thức và mẫu thức của từng phân thức
Lời giải:
\(\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{ab^2-ac^2-b^3+bc^2}=\frac{a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)}{a(b^2-c^2)-b(b^2-c^2)}\)
\(=\frac{(a^2-b^2)(b-c)-(b^2-c^2)(a-b)}{(a-b)(b^2-c^2)}=\frac{(a-b)(b-c)(a+b-b+c)}{(a-b)(b-c)(b+c)}=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}\)
\(=\frac{a-c}{b+c}\)
Ta có: \(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)}{\left(b-c\right)\left(a-b\right)\left(b+c\right)}\)
\(=\dfrac{a-c}{b+c}\)