Bài 1 :
Cho x, y, z \(\ne0\) ; A = \(\dfrac{y}{z}+\dfrac{z}{y}\) ; B = \(\dfrac{z}{x}+\dfrac{x}{z}\) ; C = \(\dfrac{x}{y}+\dfrac{y}{x}\)
Tính A\(^2\) + B\(^2\) + C\(^2\) - ABC
Bài 2 :
Cho x = \(\dfrac{a}{b+c}\) ; y = \(\dfrac{b}{c+a}\) ; z = \(\dfrac{c}{a+b}\)
Tính xy + yz + xz + 2xyz
Bài 3: Rút gọn
\(A=\left(1+\dfrac{b^2+c^2-a^2}{2abc}\right)\times\dfrac{1+\dfrac{a}{b+c}}{1-\dfrac{a}{b+c}}\times\dfrac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
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:
\(xy+yz+xz+2xyz=\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}+\frac{ac}{(b+c)(a+b)}+\frac{2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b)}{(a+b)(b+c)(c+a)}+\frac{bc(b+c)}{(a+b)(b+c)(c+a)}+\frac{ac(a+c)}{(a+b)(b+c)(c+a)}+\frac{2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b)+bc(b+c)+ca(c+a)+2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b+c)+bc(b+c+a)+ca(c+a)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+b+c)(ab+bc)+ac(a+c)}{(a+b)(b+c)(c+a)}=\frac{(c+a)b(a+b+c)+ac(a+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)[b(a+b+c)+ac]}{(a+b)(b+c)(c+a)}=\frac{(a+c)[b(a+b)+c(a+b)]}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)(b+c)(a+b)}{(a+b)(b+c)(c+a)}=1\)
Bài 3:
\(A=\frac{2bc+b^2+c^2-a^2}{2bc}\times \frac{\frac{b+c+a}{b+c}}{\frac{b+c-a}{b+c}}\times \frac{b^2+c^2-(b^2-2bc+c^2)}{a+b+c}\)
\(=\frac{(b+c)^2-a^2}{2bc}\times \frac{b+c+a}{b+c-a}\times \frac{2bc}{a+b+c}\)
\(=\frac{(b+c-a)(b+c+a)}{2bc}\times \frac{b+c+a}{b+c-a}\times \frac{2bc}{a+b+c}\)
\(=a+b+c\)
