Tham khảo

x-y-z=0

=>x=y+z và y=x-z và z=x-y

B=(1-z/x)(1-x/y)(1+y/z)+2023

\(=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{y+z}{z}+2023\)

\(=\dfrac{y}{x}\cdot\dfrac{-z}{y}\cdot\dfrac{x}{z}+2023=2023-1=2022\)

Ta có: \(x-y-z=0\)

\(\Rightarrow x-y=z\)

\(x-z=y\)

\(y+z=x\)

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

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

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

\(\Rightarrow B=-1\)

tao đẹp trai thì có gì sai

bài này mà là âm nhạc???

x-y-z=0 => x=y+z

thế vào rồi tính B

Ta có: x-y-z = 0

\(\Rightarrow\) x = y+z

\(\Rightarrow\)y = x-z

\(\Rightarrow\)z = x-y

Thay vào B ta suy ra: \(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)

= \(\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)

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

= -y/y

= -1

Vậy B = -1

x+y+z=0

nên x+y=-z; y+z=-x; x+z=-y

\(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}=-1\)

Bài này ez thôi, làm mãi rồi.

Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{xy+yz+xz}{xyz}=0\)

=> xy+yz+zx=0

=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)

Ta có: x²+2yz=x²+yz-xy-zx=(x-y)(x-z)

           y²+2xz=y²+xz-xy-yz=(x-y)(z-y)

           z²+2xy=z²+xy-yz-xz=(x-z)(y-z)

=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)