Question

[Vương Hàn]

a ) Cho 2 ( x + y ) = 5 ( y + z ) = 2 ( z + x ) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)

b ) Tìm x \(\in\) Z và x < 50 biết : M = \(\frac{\sqrt{x}-1}{2}\) có giá trị nguyên

Answer 1

[Đỗ Lê Tú Linh]

a)2(x+y)=2(z+x)

=>\(x+y=z+x\)

=>y=z

=>\(\frac{y-z}{5}=\frac{0}{5}=0\)

5(y+z)=2(z+x)

5y+5z=2z+2x

mà y=z(cmt)

nên 5y+5y-2y=2x

8y=2x

x=4y

=>\(\frac{x-y}{4}=\frac{4y-y}{4}=\frac{3y}{4}\)

=>ko thỏa mãn đề bài

Answer 2

[Nguyễn Hữu Thế]

a ) Cho 2( x + y ) = 5( y + z ) = 3( z + x ) thì x−y4=y−z5

Theo đề bài ra ta có: \(2\left(x+y\right)=5\left(y+z\right)\Rightarrow\frac{x+y}{5}=\frac{y+z}{2}\Rightarrow\frac{x+y}{15}=\frac{y+z}{6}\)

\(5\left(y+z\right)=3\left(z+x\right)\Rightarrow\frac{z+x}{5}=\frac{y+z}{3}\Rightarrow\frac{z+x}{10}=\frac{y+z}{6}\)

\(\Rightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}=\frac{x+y-y-z-z-x}{15-6-10}=\frac{0}{-1}=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+y=0\\y+z=0\\z+x=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\y=0\\z=0\end{array}\right.\)

\(\Rightarrow5x-5y=4y-4z\)(Do x,y,z=0)

\(\Rightarrow5\left(x-y\right)=4\left(y-z\right)\)

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