mk nhầm 1+c/a

Ta có : \(a+b+c\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}\left(\cdot\right)}\)

\(\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)

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

\(=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}\left(do\cdot\right)\)

\(=-1.-1.-1\)

\(=-1\)

Ta có : \(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(\Rightarrow P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)

\(\Rightarrow P+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)

\(\Rightarrow P+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{c+a}+\left(a+b+c\right).\frac{1}{a+b}\)

\(\Rightarrow P+3=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(\Rightarrow P+3=2019.10\)

\(\Rightarrow P+3=20190\)

\(\Rightarrow P=20190-3\)

\(\Rightarrow P=20187\)

Vậy P = 20187

Hình ?

Nếu \(a-b,b-c,c-a\inℤ^+\)

\(\Rightarrow\frac{|a-b|}{3}=\frac{|b-c|}{5}=\frac{|c-d|}{7}\)\(=\frac{a-b}{3}=\frac{b-c}{5}=\frac{c-a}{7}\)\(=\frac{a-b+b-c+c-a}{3+5+7}=\frac{0}{15}=0\)

\(\Rightarrow\)\(a=b=c=0\left(đpcm\right)\)

Nếu\(a-b,b-c,c-a\inℤ^-\)

\(\Rightarrow\frac{|a-b|}{3}=\frac{|b-c|}{5}=\frac{|c-d|}{7}\)\(=\frac{-a+b}{3}=\frac{-b+c}{5}=\frac{-c+a}{7}\)\(=\frac{-a+b+\left(-b\right)+c+ \left(-c\right)-a}{3+5+7}=\frac{0}{15}=0\)

\(\Rightarrow\)\(a=b=c=0\left(đpcm\right)\)

Đề sai hả bạn :V

A B C D E M N

 ( GT, KL bạn tự viết nha )

Chia hết cho 3 : 1095; 1305; 1203; 4401

Chia hết cho 5 : 1095; 1305

Chia hết cho 9 : 1305; 4401

Chia hết cho cả 3; 5; 9 là : 1305

Trả lời:

Số chia hết cho 3: 1095, 1305, 1203, 4401.

Số chia hết cho 5 : 1095, 1305.

Số chia hết cho 9: 1305, 4401.

       ~ Học tốt ~

b)Ta có: \(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow\frac{a^2}{ab}+\frac{b^2}{ab}\ge2\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge2\left(đpcm\right)\)

\(a^5-a=a\left(a^4-1\right)\)

\(=a\left(a^2+1\right)\left(a^2-1\right)\)

\(=a\left(a^2+1\right)\left(a-1\right)\left(a+1\right)\)

\(=a\left(a^2-4+5\right)\left(a-1\right)\left(a+1\right)\)

\(=a\left(a^2-4\right)\left(a-1\right)\left(a+1\right)+5a\left(a+1\right)\left(a-1\right)\)

\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a+1\right)\left(a-1\right)\)

Tích 5 số nguyên liên tiếp chia hết cho 5 nên \(a^5-a⋮5\)