\(\frac{a+b}{b+c}=\frac{c+d}{d+a}=\frac{a+b+c+d}{b+c+d+a}=1\) (dãy tỉ số bằng nhau)

\(\Rightarrow\frac{a+b}{a+c}=1\Leftrightarrow a+b=b+c\Rightarrow a=c\)(đpcm)

cảm ơn nhé

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

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

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

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

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

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

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

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

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

xin lỗi, mình nhầm chỗ này, cho mình sửa lại nha

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

Suy ra:

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

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

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

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

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

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

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

Chúc bạn học tốt

(hồi nãy mình nhầm chút xíu)

Trả lời:

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

Suy ra:

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

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

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

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

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

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

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

Chúc bạn học tốt

Ta có: 2bd = c(b + d)

=> (a + c).d = bc + cd

=> ad + cd = bc + cd

=> ad = bc

=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

Ta có : 2bd = c (b + d )

=) ( a + c ). d = bc + cd

=) ad + cd = bc + cd

=) ad = bc

=) a/b = c/ d ( đpcm)

Ta có : 2bd = c (b + d )

 => ( a + c ). d = bc + cd 

=>ad + cd = bc + cd 

=>ad = bc 

=> a/b = c/ d ( đpcm)

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

\(\Rightarrow ad< bc\)

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

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

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

a=?

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

=> a²+ab+ad+db=cb+c²+db+dc

=> a²+ab+ad+db-cb-c²-db-dc=0

=>( a²-c²) + (ab -bc) +( ad -dc)=0

=>(a+c)(a-c) +b(a-c) +d(a-c)=0

=>(a-c)(a+c+b+d)=0

=>\(\left[\begin{array}{nghiempt}a-c=0\\a+b+c+d=0\end{array}\right.\)

=>\(\left[\begin{array}{nghiempt}a=c\\a+b+c+d=0\end{array}\right.\)

đề như đệt

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

\(ad+a^2+bd+ab=bc+bd+c^2+cd\)

\(a\left(b+d\right)+a^2=c\left(b+d\right)+c^2\)

\(a+a^2=c+c^2\)

\(a=c\)

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

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

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

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

\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)

\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)

\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)

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

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

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

=>\(\left(a+b\right)\left(a+d\right)=\left(c+d\right)\left(b+c\right)\)

=> \(a^2+ab+ad+bd=c^2+bc+bd+cd\)

=>\(a^2+ab+ad-bc-c^2-cd=0\)

=>\(\left(a^2-c^2\right)+\left(ab-cd\right)+\left(ab-ac\right)=0\)

=>\(\left(a-c\right)\left(a+c\right)+d\left(a-c\right)+b\left(a-c\right)=0\)

=>\(\left(a-c\right)\left(a+b+c+d\right)=0\)

=>\(\orbr{\begin{cases}a-c=0\\a+b+c+d=0\end{cases}\left(dpcm\right)}\)

hacker 2k6

bổ sung chỗ dưới

bạn => thêm là

\(=>\orbr{\begin{cases}a=c\\a+b+c+d=0\end{cases}\left(dpcm\right)}\)