Không.

Có, khi a = 0

Có thể có phân số a/b (a, b ∈ Z, b ≠ 0) sao cho:

(m, n ∈ Z, m , n ≠ 0 , m ≠ n) khi và chỉ khi a = 0

Vì (m, n ∈ Z, m , n ≠ 0 , m ≠ n)

Ta có:

\(\frac{a}{b}< 1\\ \Rightarrow a< b\\ \Rightarrow am< bm\left(m\in N^{\cdot}\right)\\ \Rightarrow am+ab< bm+ab\\\Rightarrow a\left(b+m\right)< b\left(a+m\right)\\ \Rightarrow\frac{a}{b} < \frac{a+m}{b+m}\)

Gọi 1/4 số a là 0,25 . Ta có :

                   a . 3 - a . 0,25 = 147,07

                   a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )

                      a . 2,75 = 147,07

                         a = 147,07 : 2,75

                          a = 53,48

Theo đề ta có :

\(x=\frac{a}{m}\) \(;\)\(y=\frac{b}{m}\)

mà \(x< y\) \(\Rightarrow\frac{a}{m}< \frac{b}{m}\Rightarrow a< b\)

Có : \(x=\frac{a}{m}\Rightarrow x=\frac{2a}{2m}=\frac{a+a}{2m}\)     ; \(z=\frac{a+b}{2m}\)    và \(y=\frac{b}{m}\Rightarrow y=\frac{2b}{2m}=\frac{b+b}{2m}\) 

* Vì a < b \(\Rightarrow\) a+a < a+b \(\Rightarrow\frac{a+a}{2m}< \frac{a+b}{2m}\)\(\Rightarrow x< z\) \(\left(1\right)\)

* Vì \(a< b\)\(\Rightarrow a+b< b+b\Rightarrow\frac{a+b}{2m}< \frac{b+b}{2m}\Rightarrow z< y\)\(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) nên ta có :

\(\frac{a+a}{2m}< \frac{a+b}{2m}< \frac{b+b}{2m}\Rightarrow x< z< y\) \(\left(đpcm\right)\)

mk không hiểu

đề đúng mà bn

đề đúng thì giải giùm ik bạn ơi

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)

Ta có: \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\frac{ak\cdot bk\cdot ck\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\cdot\left(ak+bk\right)\cdot\left(bk+ck\right)\cdot\left(ck+ak\right)}\)

\(=\frac{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)

Vậy: H=1

đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)

theo giả thiết ta có \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

thay \(H=\frac{ak.bk.ck\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(ak+bk\right)\left(bk+ck\right)\left(ck+ak\right)}\)

\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left[k\left(a+b\right)\right]\left[k\left(b+c\right)\right]\left[k\left(c+a\right)\right]}\)

\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc.k\left(a+b\right).k\left(b+c\right).k\left(c+a\right)}\)

\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)

Vậy H = 1