Sửa đề: Chứng minh a+d=b+c

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

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

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

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

=>\(\left(d+c\right)\cdot\left(2a+d-c\right)=\left(c+d\right)\cdot\left(2b+c-d\right)\)

=>2a+d-c=2b+c-d

=>2a-2b=c-d-d+c=2c-2d

=>a-b=c-d

=>a+d=b+c

a) (a + b + c + d)(a - b - c - d)

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

= (aa + ab + ac + ad) - (ba + bb + bc + bd) - (ca + cb + cc + cd) - (da + db + dc + dd)

= aa - bb - cc - dd

\(ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1;\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d};\frac{a-b}{b}=\frac{c-d}{d}\)

a) \(\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)(đúng)

b)\(\Leftrightarrow\frac{a}{b}-1=\frac{c}{d}-1\Leftrightarrow\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)(đúng)

a)Do b,d>0

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

b)Do b,d>0

=>\(ad>bc\Leftrightarrow\frac{ad}{bd}>\frac{bc}{bd}\Rightarrow\frac{a}{b}>\frac{c}{d}\)

a,

b,  a/b < c/d => ad < cb
=>ad +ab < bc+ab
=> a(d+b) < b(a+c)
=> a/b < a+c/d+b (1)
* a/b < c/d => ad<cb
=> ad + cd < cb +cd
=> d(a+c) < c(b+d) 
=> c/d > a+c/b+d (2)
Từ (1) và (2) => a/b < a+c/b+d < c/d

Vì \(b,d>0\)nên \(bd>0\)

Ta có:  \(\frac{a}{b}< \frac{c}{d}\)

\(\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\)

\(\Leftrightarrow ad< bc\)vì \(bd>0\)