a, \(\frac{a}{b}=\frac{ad}{bd};\frac{c}{d}=\frac{bc}{bd}\)

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

b, Theo câu a ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\Rightarrow ad+ab< bc+ab\Rightarrow a\left(b+d\right)< b\left(a+c\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)

Lại có: \(ad< bc\Rightarrow ad+cd< bc+cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)

Từ (1) và (2) => đpcm

a, \(\frac{a}{b}=\frac{ad}{bd};\frac{c}{d}=\frac{bc}{bd}\)

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

b, Theo câu a, ta có:

\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\Rightarrow ad+ab< bc+ab\Rightarrow a\left(b+d\right)< b\left(a+c\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\)(1)

Lại có: \(ad< bc\Rightarrow ad+cd< bc+cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\)(2)

Từ (1) và (2) => đpcm.

Còn nha. Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}^{\left(1\right)}\)

Lại có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}^{\left(2\right)}\)

Từ (1) và (2) => đpcm

Ta có : \(b>0,d>0,\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)                                                                         ( 1 )

\(\Rightarrow ad+ab< bc+ab\)

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

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

Vì \(b>0,d>0,\frac{a}{b}< \frac{c}{d}\)

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

\(\Rightarrow ad+cd< bc+cd\)                                                             ( 2 )

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

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

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

Ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\Rightarrow ad.ab< bc.ab\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\)(1)

\(ad< bc\Rightarrow ad.cd< bc.cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\)(2)

Từ (1) và (2) ta được: \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

a) \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\) (quy đồng mẫu chung)

Vì b,d > 0 nên bd > 0. Do đó ad < bc (đpcm)

b) ad < bc \(\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\) (cùng chia cho bd)

Vì b,d > 0 nên bd > 0. Do đó \(\frac{a}{b}< \frac{c}{d}\) (rút gọn tử và mẫu)

a, Ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{bd}< \frac{cb}{db}\Rightarrow ad< cb\) 

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

Ta có

a/b<c/d \(\Leftrightarrow\)ad<bc           (1)

Thêm ab vào 2 vế của (1) ta được:

ad+ab<bc+ab hay a(b+d)<b(a+c) =>a/b<a+c/b+d            (2)

Thêm cd vào 2 vế của (2) ta được:

ad+cd<bc+cd  hay d(a+c)<c(b+d) =>c/d>a+c/b+d            (3)

Từ (2) và (3) suy ra:a/b<a+c/b+d<c/d

**** bạn

a, \(\frac{a}{b}=\frac{a\left(b+d\right)}{b\left(b+d\right)}=\frac{ab+ad}{b\left(b+d\right)}\)

 \(\frac{a+c}{b+d}=\frac{b\left(a+c\right)}{b\left(b+d\right)}=\frac{ab+bc}{b\left(b+d\right)}\)

VCif a/b<c/d => ad<bc

=> ab + ad < ab +ad

=> a/b < (a+c) / (b+d) (1)

Cm tương tự :

(a+c) / (b+d) < c/d (2)

 Từ 1 và 2 => DPCM

Ta có:\(\frac{3a+b+c+d}{a}=\frac{a+3b+c+d}{b}=\frac{a+b+3c+d}{c}=\frac{a+b+c+3d}{d}\)

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

\(\Rightarrow\orbr{\begin{cases}a+b+c+d=0\\a=b=c=d\end{cases}}\)

\(TH1:a+b+c+d=0\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)

\(\Rightarrow Q=\left(\frac{-\left(c+d\right)}{c+d}\right)^2+\left(\frac{-\left(a+d\right)}{a+d}\right)^2+\left(\frac{c+d}{-\left(c+d\right)}\right)^2+\left(\frac{a+d}{-\left(a+d\right)}\right)^2\)

\(\Rightarrow Q=\left(-1\right)^2\cdot4=1\cdot4=4\)

\(TH2:a=b=c=d\)

\(\Rightarrow Q=\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2=1^2\cdot4=1\cdot4=4\)

Vậy Q=4

ta có:

\(\frac{3}{a+b}+\frac{2}{c+d}+\frac{a+b}{\left(a+c\right)\left(b+d\right)}\ge\frac{3}{a+b}+\frac{2}{c+d}+\frac{4\left(a+b\right)}{\left(a+b+c+d\right)^2}\)

xét hiệu:

\(\frac{3}{a+b}+\frac{2}{c+d}+\frac{4\left(a+b\right)}{\left(a+b+c+d\right)^2}-\frac{12}{a+b+c+d}\)

\(=\frac{3}{a+b}+\frac{2}{c+d}-\frac{8\left(a+b\right)+12\left(c+d\right)}{\left(a+b+c+d\right)^2}\)

đặt a+b=x;c+d=y

\(\Rightarrow\frac{3}{a+b}+\frac{2}{c+d}-\frac{8\left(a+b\right)+12\left(c+d\right)}{\left(a+b+c+d\right)^2}=\frac{3}{x}+\frac{2}{y}-\frac{8x+12y}{\left(x+y\right)^2}\ge\frac{3}{x}+\frac{2}{y}-\frac{8x+12y}{4xy}=\frac{3}{x}+\frac{2}{y}-\frac{2}{y}-\frac{3}{x}=0\)

\(\Rightarrow\frac{3}{a+b}+\frac{2}{c+d}+\frac{4\left(a+b\right)}{\left(a+b+c+d\right)^2}\ge\frac{12}{a+b+c+d}\)

\(\Rightarrow\frac{3}{a+b}+\frac{2}{c+d}+\frac{a+b}{\left(a+c\right)\left(b+d\right)}\ge\frac{12}{a+b+c+d}\)

=>đpcm

dấu "=" xảy ra khi a=b=c=d