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Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
\(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)
\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)
\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)
Bài 2:
a)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=> a = b = c
b)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
=> x = y = z (theo a)
Thay x = y = z vào biểu thức, ta có:
\(M=\dfrac{x^{333}.x^{666}}{x^{999}}=1\)
c)
\(ac=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(ab=c^2\Rightarrow\dfrac{b}{c}=\dfrac{c}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Rightarrow a=b=c\)
Thay a = b = c vào biểu thức, ta có:
\(M=\dfrac{a^{333}}{a^{111}.a^{222}}=1\)
Ta có \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c=\dfrac{a+b+c}{2}\)
Suy ra
M = (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) + x2
= x2 - ax - bx + ab + x2 - bx - cx + bc + x2 - ax - cx + ac + x2
= 4x2 - 2ax - 2bx - 2cx + ab + bc + ac
= (2x)2 - 2x(a + b + c) + ab + bc + ac
= \(\left(2\cdot\dfrac{a+b+c}{2}\right)^2-\left(2\cdot\dfrac{a+b+c}{2}\right)\left(a+b+c\right)+ab+bc+ac\)
= ab + bc + ac
Theo tính chất của dãy tỉ số bằng nhau:
\(x=\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{b+c+a+c+a+b}{a+b+c}=2\)
⇒ A = (x2 - x + 1)10 = (22 - 2 + 1)10 = 310