Lớp 6 hả

mk ko hiểu ?

Toán lớp 6 mà khó quá????? 

Thay \(a+b+c\) vào \(A\) ta được:

\(A=\frac{a}{2017-c}+\frac{b}{2017-a}+\frac{c}{2017-b}\)

\(=\frac{a}{a+b+c-c}+\frac{b}{a+b+c-a}+\frac{c}{a+b+c-b}\)

\(=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}\)

Ta có:

\(\frac{a}{a+b}< \frac{a+b}{a+b+c}\)

\(\frac{b}{b+c}< \frac{b+a}{a+b+c}\)

\(\frac{c}{c+a}< \frac{c+b}{a+b+c}\)

Cộng vế với vế ta được:

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}\)\(=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow A< 2\left(1\right)\)

Lại có:

\(\frac{a}{a+b}>\frac{a}{a+b+c}\)

\(\frac{b}{b+c}>\frac{b}{a+b+c}\)

\(\frac{c}{c+a}>\frac{c}{a+b+c}\)

Cộng vế với vế ta lại được:

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)\(=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow A>1\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow1< A< 2\)

Vậy \(A\) không phải là số nguyên (Đpcm)

cái này chứng minh 1 < A < 2. mình chỉ bít chứng minh 1 < A thui 

Ta có \(\frac{a}{2017-c}>\frac{a}{2017};\frac{b}{2017-a}>\frac{b}{2017};\frac{c}{2017-b}>\frac{c}{2017}\) 

suy ra \(A>\frac{a}{2017}+\frac{b}{2017}+\frac{c}{2017}=\frac{2017}{2017}=1\)

=> A > 1

Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=@\Leftrightarrow\left\{{}\begin{matrix}a=b@\\c=d@\end{matrix}\right.\)

khi đó: \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{b^{2017}@^{2017}+b^{2017}}{d^{2017}@^{2017}+d^{2017}}=\dfrac{b^{2017}\left(@^{2017}+1\right)}{d^{2017}\left(@^{2017}+1\right)}=\dfrac{b^{2017}}{d^{2017}}\)

\(\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\dfrac{\left(b@-b\right)^{2017}}{\left(d@-d\right)^{2017}}=\dfrac{\left[b\left(@-1\right)\right]^{2017}}{\left[d\left(@-1\right)\right]^{2017}}=\dfrac{b^{2017}}{d^{2017}}\)

Ta có điều phải chứng minh

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{b^{2017}\cdot k^{2017}+d^{2017}\cdot k^{2017}}{b^{2017}+d^{2017}}=k^{2017}\)

\(\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}=\dfrac{\left(bk+dk\right)^{2017}}{\left(b+d\right)^{2017}}=k^{2017}\)

Do đó: \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)

vì \(\frac{a}{b}\)=\(\frac{c}{d}\)=>\(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\) 

áp dụng tính chất dãy tỉ số bằng nhau

=> \(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\)= \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\)=\(\frac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}\)=\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)(diều phải chứng minh

Từ \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra a=bk

           c=dk

Ta có

\(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(bk\right)^{2017}+b^{2017}}{\left(dk\right)^{2017}+d^{2017}}=\frac{b^{2017}.k^{2017}+b^{2017}}{d^{2017}.k^{2017}+d^{2017}}=\frac{b^{^{2017}}\left(k^{2017}+\right)}{d^{2017}\left(k^{2017}+1\right)}=\frac{b^{2017}}{d^{2017}}\)(1)

Ta có

\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\frac{\left(bk-b\right)^{2017}}{\left(dk-d\right)^{2017}}=\frac{\left(b\left(k-1\right)\right)^{2017}}{\left(d\left(k-1\right)\right)^{2017}}=^{\frac{b^{2017}}{d^{2017}}}\)(2)

Từ (1) và (2)

Ta suy ra

\(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)

từ gt: \(\frac{a}{b}\)=\(\frac{c}{d}\)suy ra ad=bc

\(\frac{a^{2017}+b^{2017}=\left(a-b\right)^{2017}}{^{c^{2017}}+d^{2017}=\left(c-d\right)^{2017}}\)

suy ra \(a^{2017}+b^{2017}.\left(c-d\right)^{2017}=c^{2017}+d^{2017}.\left(a-b\right)^{2017}\)

\(a^{2017}+b^{2017}.c^{2017}-b^{2017}.d^{2017}=c^{2017}+d^{2017}.a^{2017}-d^{2017}.b^{2017}\)

theo mình nghĩ là\(b^{2017}.c^{2017}=d^{2017}.a^{2017}\)

bc=da