Question

Cho \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\).Tính giá trị biểu thức \(A=\frac{x-y+z}{x+2y-z}\)

Answer 1

+ \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=\frac{x-y+z}{2-5+7}=\frac{x-y+x}{4}\Rightarrow x-y+x=2x\)

+ \(\frac{x}{2}=\frac{2y}{10}=\frac{z}{7}=\frac{x+2y-z}{2+10-7}=\frac{x+2y-z}{5}\Rightarrow x+2y-z=\frac{5x}{2}\)

\(\Rightarrow\frac{x-y+z}{x+2y-z}=\frac{2x.2}{5x}=\frac{4}{5}\)

Answer 2

4/5 bạn nhé

Answer 3

  

+ x2 =y5 =z7 =x−y+z2−5+7 =x−y+x4 ⇒x−y+x=2x

+ x2 =2y10 =z7 =x+2y−z2+10−7 =x+2y−z5 ⇒x+2y−z=5x2 

⇒x−y+zx+2y−z =2x.25x =45 

Đảm bảo đúng  

Answer 4

4/5 nhé

Answer 5

4/5 nha kb với mk nhé

Answer 6

=4/5 nhé

Answer 7

trả lời

  4/5

 hok tốt!