a) a/b=ad/bd

c/d=cb/db

mà a/b<c/d=>ad/bd<cb/bd=>ad<bc

b)ad<bc=>ad/bd<bc/bd=> a/b<c/d

Ta có:

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

=>ad+ab<bc+ab

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

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

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

=>ad+cd<bc+cd

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

=>\(\frac{a+c}{b+d}< \frac{c}{d}\) (2)

Từ (1) và (2) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)(đpcm)

---------------

\(\frac{-1}{3}=\frac{-8}{24}>\frac{-9}{24}>\frac{-10}{24}>\frac{-11}{24}>\frac{-12}{24}=\frac{-1}{2}\)

---------------

\(\frac{-1}{5}< \frac{-1}{4}< \frac{-1}{3}< \frac{-1}{2}< -1< 0< \frac{1}{5}\)

\(\frac{-1}{2}=\frac{\left(-1\right).12}{2.12}=\frac{-12}{24}\)

\(\frac{-1}{3}=\frac{\left(-1\right).8}{3.8}=\frac{-8}{24}\)

\(\frac{-8}{24}< x< \frac{-12}{24}\)

\(\Rightarrow x=\left\{\frac{-9}{24};\frac{-10}{24};\frac{-11}{24}\right\}\)

Xét hiệu:       \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)\(=0\)   (do  a+b+c = 0)

\(\Rightarrow\)\(a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\)\(a^3+b^3+c^3=3abc\)  (đpcm)

a) do a/b>c/d (b>0,d>0)

=> ad>bc => ad+ab>bc+ab

a.(d+b)>b(c+a)   => a/b=c+a/b+d (1)

tương tự cộng với cd là xong

b) 1/3<15/48,14/48,13/68<1/4

Mọi người ơi giúp e với :>>

đang cần gấp mn ơi:>>>

Bài 1 :

\(a,\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)

Ta có : \(VT=\left(a-b\right)+\left(c-d\right)-\left(a-c\right)\)

                 \(=a-b+c-d-a+c\)

                 \(=-\left(b+d\right)=VP\)

\(\Rightarrow\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)

\(b,\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)

Ta có : \(VT=\left(a-b\right)-\left(c-d\right)+\left(b+c\right)\)

                 \(=a-b-c+d+b+c\)

                 \(=a+d=VP\)

\(\Rightarrow\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)

Sao k ai giúp mik v, huhu?

tại em mới lớp 6 thôi