Áp dụng tích chất dãy tỉ số bằng nhau ta có :

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}=\dfrac{x+y+z}{x+y+z}=1\\ \Rightarrow\left\{{}\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\)

\(\Rightarrow\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{x+y}{y}.\dfrac{y+z}{z}.\dfrac{x+z}{x}=\dfrac{2z}{y}.\dfrac{2x}{z}.\dfrac{2y}{x}=8\)

Vào đây nhé: Câu hỏi của Vũ Ngọc Minh Anh - Toán lớp 7 | Học trực tuyến

a, H = \(2^{2010}-2^{2009}-2^{2008}-...-2-1\)

\(\Leftrightarrow\) 2H = \(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)

\(\Leftrightarrow\) 2H - H = \((2^{2011}-2^{2010}-2^{2009}-...-2^2-2)\) - \((2^{2010}-2^{2009}-2^{2008}-...-2-1)\)

\(\Leftrightarrow\) H = \(2^{2011}-2.2^{2010}+1\)

\(\Leftrightarrow\) H = \(2^{2011}-2^{2011}+1\)

\(\Leftrightarrow\) H = 1

Vậy H = 1

a)H=2²⁰¹⁰-2²⁰⁰⁹-...-2-1

=>2H=2(2²⁰¹⁰-2²⁰⁰⁹-...-2-1)

=>2H=2²⁰¹¹-2²⁰¹⁰-...-2²-2

=>2H-H=(2²⁰¹¹-2²⁰¹⁰-...-2²-2)-(2²⁰¹⁰-2²⁰⁰⁹-...-2-1)

=>H=2²⁰¹¹-1

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)

\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)

\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)

\(\circledast\) Với \(x=y=z\) thì \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

\(\circledast\) Với \(x+y+z=0\) thì\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)

Khi đó \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)

Ta có :

\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}\\ \Leftrightarrow\dfrac{x+y+z}{z}=\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\left(cùngcộngthêm2\right)\)

TH1: \(x+y+z\ne0\)

\(\Rightarrow x=y=z\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2=8\)

TH2: \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)(*)

\(\Rightarrow P=\left(1+\dfrac{-\left(y+z\right)}{y}\right)\left(1+\dfrac{-\left(z+x\right)}{z}\right)\left(1+\dfrac{-\left(x+y\right)}{z}\right)\\ =\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{z}\right)\\ =\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{z}\right)\\ =-1\)

Vậy P=8 hoặc P=-1

Ta có : từ x - y - z =0

\(\Rightarrow x-z=y\) ; \(-z=y-x\) ; \(y+z=x\)

Lại có \(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)

\(\Rightarrow B=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{y+z}{z}\)

thay các hằng đẳng thức vừa tìm được vào B

\(\Rightarrow B=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)

vậy B = -1

tik mik nha !!!

ta có x-y-z=0

->x=y+z

y=x-z

z=x-y

B=\(\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1-\dfrac{y}{z}\right)\)

B=\(\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)

B=\(\dfrac{y}{x}.\left(-\dfrac{z}{y}\right)\left(\dfrac{x}{z}\right)\)

B=\(\dfrac{-\left(xyz\right)}{xyz}\)

B=-1