Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

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

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8

Lời giải:
Ta có:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2023.\frac{2024}{2023}$

$\Leftrightarrow 1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=2024$

$\Leftrightarrow 3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}=2024$

$\Leftrightarrow 3+B=2024$

$\Leftrightarrow B=2021$

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

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

B=\(\left(x+y\right)\left(y+z\right)\left(z+x\right).M=0\)

M ở đâu ra thế bạn

Phạm Văn Chí

1 nha bạn

k tui nha 

thank

x-y-z=0

=> x=y+z

     y=x-z

    -z=y-x

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

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

B=(y/x)(-z/y)(x/z)

B=(-z.y.x)/(x.y.z)

B=-1

thank ban nha

B=-1    ^_^

y x 8,01 - y : 100 = 38
y x 8,01 - y x 0,01 = 38
y x ( 8,01 - 0,01 ) = 38
y x 8 = 38
y = 38 : 8

mk chắc chắn 

p/s tham khảo nhé ^_^

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