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\(\frac{x^2-3x+2}{x^3-1}=\frac{x^2-2x-x+2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x.\left(x-2\right)-\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{\left(x-1\right).\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x-2}{x^2+x+1}\)
c) \(\frac{a\left(a^2-ab+b^2\right)}{b\left(a+b\right)\left(a^2-ab+b^2\right)}\)
=\(\frac{a}{b\left(a+b\right)}\)
Bài 1:
Đặt \(\left(\frac{x}{y}; \frac{y}{z}; \frac{z}{x}\right)=(a,b,c)\Rightarrow abc=1\)
Khi đó:
\(A^2+B^2+C^2-ABC=(b+\frac{1}{b})^2+(c+\frac{1}{c})^2+(a+\frac{1}{a})^2-(a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})\)
\(=b^2+\frac{1}{b^2}+2+c^2+\frac{1}{c^2}+2+a^2+\frac{1}{a^2}+2-(ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab})(c+\frac{1}{c})\)
\(a^2+b^2+c^2+(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})+6-[abc+\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)+\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)+\frac{1}{abc}]\)
\(=a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+6-[1+\left(\frac{abc}{c^2}+\frac{abc}{a^2}+\frac{abc}{b^2}\right)+\left(\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}\right)+1]\)
\(=a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+6-[1+(\frac{1}{c^2}+\frac{1}{b^2}+\frac{1}{a^2})+(a^2+b^2+c^2)+1]\)
\(=4\)
Câu 2:
Ta có:
\(xy+yz+xz+2xyz=\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}+\frac{ac}{(b+c)(a+b)}+\frac{2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b)}{(a+b)(b+c)(c+a)}+\frac{bc(b+c)}{(a+b)(b+c)(c+a)}+\frac{ac(a+c)}{(a+b)(b+c)(c+a)}+\frac{2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b)+bc(b+c)+ca(c+a)+2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b+c)+bc(b+c+a)+ca(c+a)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+b+c)(ab+bc)+ac(a+c)}{(a+b)(b+c)(c+a)}=\frac{(c+a)b(a+b+c)+ac(a+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)[b(a+b+c)+ac]}{(a+b)(b+c)(c+a)}=\frac{(a+c)[b(a+b)+c(a+b)]}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)(b+c)(a+b)}{(a+b)(b+c)(c+a)}=1\)
Câu 1:
\(\text{a) }\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)
\(\text{b) }\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\\ =\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\\ =\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)^2}{5b\left(x-1\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\\ =-\dfrac{2ax-2a}{5bx+5b}\)
\(\text{c) }\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\text{d) }\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
\(\text{e) }\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x+y\right)^3}\\ =\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\\ =\dfrac{x^3+y^3}{x^4-xy^3}\)
Câu 3:
\(\text{ a) }\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
\(\text{b) }\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\\ =\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\\ =\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\\ =\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}\\ =\dfrac{a+b-c}{a-b+c}\)
\(\text{c) }\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\\ =\dfrac{2x^3-x^2-6x^2+3x-15x+45}{3x^3-10x^2-9x^2+3x+30x-9}\\ =\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\\ =\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-6x+5x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\\ =\dfrac{\left(x-3\right)\left[\left(2x^2-6x\right)+\left(5x-15\right)\right]}{\left(x-3\right)\left[\left(3x^2-9x\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left[x\left(x-3\right)+5\left(x-3\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-3\right)\left(3x-1\right)}\\ =\dfrac{x+5}{3x-1}\)
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
a)
\(\dfrac{x^2+x-6}{x^3-4x^2-18x+9}=\dfrac{x^2+3x-2x-6}{x^3+3x^2-7x^2-21x+3x+9}\)
\(=\dfrac{x\left(x+3\right)-2\left(x+3\right)}{x^2\left(x+3\right)-7x\left(x+3\right)+3\left(x+3\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x^2-7x+3\right)\left(x+3\right)}=\dfrac{x-2}{x^2-7x+3}\)
a: \(=\dfrac{x\left(y-1\right)-\left(y-1\right)}{y\left(1-z\right)-\left(1-z\right)}=\dfrac{\left(y-1\right)\left(x-1\right)}{\left(1-z\right)\left(y-1\right)}=\dfrac{x-1}{1-z}\)
b: \(=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a-b\right)\left(a+b\right)-\left(a+b\right)}=\dfrac{a-b}{a-b-1}\)
c: \(=\dfrac{\left(a+1\right)\left(a^2-a+1\right)}{2\left(a+1\right)^2}=\dfrac{a^2-a+1}{2a+2}\)