Vì a/b<c/d nên a.d<c.b

=>2018.a.d<2018.c.b

=>2018.a.d+c.d<2018.c.b+c.d

=>2018a+c/2018b+d<c/d

Vậy ta đã chứng minh 2018a+c/2018b+d<c/d.

Vì a/b<c/d nên a.d<c.b

=>2018.a.d<2018.c.b

=>2018.a.d+c.d<2018.c.b+c.d

=>2018a+c/2018b+d<c/d

Vậy ta đã chứng minh 2018a+c/2018b+d<c/d.

Vì a/b<c/d nên a.d<c.b

. =>2018.a.d<2018.c.b.

=>2018.a.d+c.d<2018.c.b+c.d.

=>2018a+c/2018b+d<c/d.

Vậy ta đã chứng minh được 2018a+c\2018b+d<c\d

Vì a/b < c/d (Với a,b,c,d thuộc N*)

=> ad<bc

=>  2018ad < 2018bc

=> 2018ad + cd < 2018bc +cd

=> (2018a + c).d < (2018b+d).c

=> 2018a +c / 2018b + d < c/d

lho1 quá ,bỏ qua nhé bạn!

\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)

\(\Rightarrow2019ad< 2019bc\)

\(\Rightarrow2019ad+cd< 2019bc+cd\)

\(\Rightarrow d\left(2019a+c\right)< c\left(2019b+d\right)\)

\(\Rightarrow\frac{2019a+c}{2019b+d}< \frac{c}{d}\left(đpcm\right)\)

Ta có \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)

\(\Leftrightarrow2018ad< 2018bc\)

\(\Leftrightarrow2018ad+cd< 2018bc+cd\)

\(\Leftrightarrow d\left(2018a+c\right)< c\left(2018b+d\right)\)

\(\Rightarrow\frac{2018a+c}{2018b+d}< \frac{c}{d}\left(đpcm\right)\)

1a ) N= (-a-b)-(-c+d)-(-a+c)

=> N=-a-b+c-d+a-c

=>N=(-a+a)-b+(c-c)-d

=>N= -b-d

b) thay b=11 , d=-7 vào N ta có :

N=-11-(-7)

=>N= -4

c) thay b=-2 , d=-5 vào N ta có :

N= -2-(-5)

N=3

vì a<0

nên -2018.-a = 2018a => 2018a > 0 (1)

vì b>0

nên -2018.b = -2018b => -2018b < 0 (2)

từ (1) và (2) ta có : -2018b<2018a

vậy -2018a >-2018b

Vì a/b<c/d nên a.d<c.b

=>2018.a.d<2018.c.b

=>2018.a.d+c.d<2018.c.b+c.d

=>2018a+c/2018b+d<c/d

Vậy ta đã chứng minh 2018a+c/2018b+d<c/d.

\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)

\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)

\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)

\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)

\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)

\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)

\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)

\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)

\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)