Question

 

Answer 1

giúp mình với 

Answer 2

Bài 2:

a) Áp dụng TCDTSBN ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{30}{10}=3\)

\(\dfrac{x}{2}=3\Rightarrow x=6\\ \dfrac{y}{3}=3\Rightarrow y=9\\ \dfrac{z}{5}=3\Rightarrow z=15\)

b) Áp dụng TCDTSBN ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2.3+3.5}=\dfrac{22}{11}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\\ \dfrac{y}{3}=2\Rightarrow y=6\\ \dfrac{z}{5}=2\Rightarrow z=10\)

 

Answer 3

Bài 2:
a, Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{30}{10}=3\)
* \(\dfrac{x}{2}=3\Rightarrow x=6\)
* \(\dfrac{y}{3}=3\Rightarrow y=9\)
* \(\dfrac{z}{5}=3\Rightarrow z=15\)
b, Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{2y}{6}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{2-6+15}=\dfrac{22}{11}=2\)
* \(\dfrac{x}{2}=2\Rightarrow x=4\)
* tương tự như tìm x để tính y, z nhé.
 

Answer 4

 

H này ms chụp :v