Question

Tính \(\dfrac{xyz}{x+y+z}\) biết \(\left(x+y\right):\left(8-z\right):\left(y+z\right):\left(10+z\right)=2:5:3:4\)

Answer 1

dài lắm

\(\left(x+y\right):\left(8-z\right):\left(y+z\right):\left(10+z\right)=2:5:3:4\\ < =>\dfrac{x+y}{2}=\dfrac{8-z}{5}=\dfrac{y+z}{3}=\dfrac{10-z}{4}\left(1\right)\)

\(\left(1\right)=>\dfrac{8-z}{5}=\dfrac{10+z}{4}\\ < =>4\left(8-z\right)=5\left(10+z\right)\\ < =>32-4z=50+5z\\ < =>-9z=18\\ < =>z=-2\left(2\right)\)

\(\left(1\right)=>\dfrac{y+z}{3}=\dfrac{8-z}{5}\left(3\right)\)

thay (2) vào (3)

\(=>\dfrac{y-2}{3}=\dfrac{8+2}{5}\\ < =>\dfrac{y-2}{3}=2\\ < =>y=8\left(4\right)\)

\(\left(1\right)=>\dfrac{x+y}{2}=\dfrac{8-z}{5}\left(5\right)\)

thay 4 và 2 vào 5

\(=>\dfrac{x+8}{2}=\dfrac{8+2}{5}\\ < =>\dfrac{x+8}{2}=2\\ < =>x=-4\left(6\right)\)

\(=>\dfrac{xyz}{x+y+z}\\ =\dfrac{\left(-2\right).8.\left(-4\right)}{\left(-4\right)+8+\left(-2\right)}\\ =\dfrac{64}{2}\\ =32\)

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