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a/ x3 + x2 z + y2 z - xyz + y3
= (x + y)(x2 - xy + y2) + z(x2 - xy + y2)
= (x2 - xy + y2)(x + y + z)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
* 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) \(\left(-5a^4\right)\cdot\left(a^2b-ab^2\right)\)
\(=\left(-5a^4\cdot a^2b\right)-\left(-5a^4\cdot ab^2\right)\)
\(=-5a^6b+5a^5b^2\)
b) \(\left(x+2y\right)\left(xy^2-2y^3\right)\)
\(=x^2y^2-2xy^3+2xy^3-4y^4\)
\(=x^2y^2-4y^4\)
`a, (-5a^4)(a^2b - ab^2)`
`= -5(a^(4+2) . b) + 5a^(4+1) . b^2`
`= -5a^6b + 5a^5b^2`
`b, (x+2y)(xy^2-2y^3)`
`= x^2y^2 + 2xy^3 - 2xy^3 - 4y^4`
a, Với y >= 0
hpt có dạng \(\left\{{}\begin{matrix}2x+y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=9\\y=x-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)(ktmđk)
Với y < 0 hpt có dạng
\(\left\{{}\begin{matrix}2x-y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3-6=-9\end{matrix}\right.\)(tm)
b, bạn tự làm
c, đk : x>= 3
\(\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\\sqrt{x+3}-3\left|y-2\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\2\sqrt{x+3}-6\left|y-2\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\left|y-2\right|=1\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y-2=\dfrac{1}{7}\\y-2=-\dfrac{1}{7}\end{matrix}\right.\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\)
bạn tự giải nốt nhé
a, (y-3)(y+3)=y2-32=y2-9 (hằng đẳng thức)
b, (a-b-c)2 - (a-b+c)2= ((a-b-c)-(a-b+c)).((a-b-c)+(a-b+c))
=(a-b-c-a+b-c).(a-b-c+a-b+c)=-2c+2a-2b
c, (m+n)(m2 -mn+n2)=m3+n3(hằng đẳng thức)
d
mình bận hồi mình làm tiếp