Cho xyz=1.Tính M=x/xy+x+1 + y/yz+y+1 + z/zx+z+1 21/09/2021 Bởi Kaylee Cho xyz=1.Tính M=x/xy+x+1 + y/yz+y+1 + z/zx+z+1
Đáp án: bằng 1 đó bạn à Giải thích các bước giải: \(M=\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\) \(=\dfrac{x}{xyz+xy+x}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\) \(=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\) \(=\dfrac{y+1}{yz+y+1}+\dfrac{z}{xz+z+1}=\dfrac{xyz+y}{xyz+yz+y}+\dfrac{z}{xz+z+1}\) \(=\dfrac{xz+1}{xz+z+1}+\dfrac{z}{xz+z+1}=\dfrac{xz+z+1}{xz+z+1}=1\) Bình luận
$\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{zx+z+1}$ $=\dfrac{x}{xy+x+xyz}+\dfrac{y}{yz+y+1}+\dfrac{z}{zx+z+xyz}$ (do $xyz=1$) $=\dfrac{x}{x(yz+y+1)}+\dfrac{y}{yz+y+1}+\dfrac{z}{z(x+1+xy)}$ $=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{1}{x+1+xy}$ $=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{xyz}{x+xyz+xy}$ Vì $xyz=1$ $=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{xyz}{x(1+yz+y)}$ $=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{yz}{1+yz+y}$ $=\dfrac{yz+y+1}{yz+y+1}=1$ Bình luận
Đáp án: bằng 1 đó bạn à
Giải thích các bước giải:
\(M=\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{x}{xyz+xy+x}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{y+1}{yz+y+1}+\dfrac{z}{xz+z+1}=\dfrac{xyz+y}{xyz+yz+y}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{xz+1}{xz+z+1}+\dfrac{z}{xz+z+1}=\dfrac{xz+z+1}{xz+z+1}=1\)
$\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{zx+z+1}$
$=\dfrac{x}{xy+x+xyz}+\dfrac{y}{yz+y+1}+\dfrac{z}{zx+z+xyz}$ (do $xyz=1$)
$=\dfrac{x}{x(yz+y+1)}+\dfrac{y}{yz+y+1}+\dfrac{z}{z(x+1+xy)}$
$=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{1}{x+1+xy}$
$=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{xyz}{x+xyz+xy}$ Vì $xyz=1$
$=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{xyz}{x(1+yz+y)}$
$=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{yz}{1+yz+y}$
$=\dfrac{yz+y+1}{yz+y+1}=1$