Ta có :\(\text{VT = A + B}\)

\(\text{= ( a + b + 5 ) + ( b – c – 9 )}\)

\(\text{= a + b + 5 + b – c – 9}\)

\(\text{= a + ( b + b ) – c + ( 5 – 9 )}\)

\(\text{= a + 2b – c – 4 (1)}\)

\(\text{VP = C – D}\)

\(\text{= ( b – c – 4 ) – ( -b – a )}\)

\(\text{= b – c – 4 + b + a}\)

\(\text{= ( b + b ) – c + a – 4}\)

\(\text{= 2b – c + a – 4}\)

\(\text{= a + 2b – c – 4 (2)}\)

\(\text{từ (1) và (2) suy ra}\)\(\text{ A + B = C – D ( đpcm ) }\)

a)a/b=c/d=a+b/c+d=a-b/c-d(tc day ti so bang nhau)

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

b)a/b=c/d=>5a/5b=2c/2d=5a+2c/5c+2d(*) va a/b=4c/4d=a-4c/c-4d(**)

c)a/b=c/d=a+b/c+d=>(a/b)^2=ab/cd=(a+b/c+d)^2

1) Ta có:
\(\dfrac{a}{a+b}\)=\(\dfrac{c}{c+d}\)
=>a.(c+d) = c.(a+b)
a.c+a.d = a.c+b.d
Do đó a.d=b.d
=>\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)( đpcm)

Câu 2: 

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+2c}{3b+2d}=\dfrac{3bk+2dk}{3b+2d}=k\)

\(\dfrac{-5a+3c}{-5b+3d}=\dfrac{-5bk+3dk}{-5b+3d}=k\)

=>\(\dfrac{3a+2c}{3b+2d}=\dfrac{-5a+3c}{-5b+3d}\)

b: \(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)

\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{c\left(2c-a\right)}{d\left(2d-b\right)}=\dfrac{dk}{d}\cdot\dfrac{2dk-bk}{2d-b}=k^2\)

=>\(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)

Ta sẽ lần lượt chứng minh:\(\frac{a}{b}\)<\(\frac{5a+2c}{5b+2d}\)và \(\frac{5a+2c}{5b+2d}\)<\(\frac{c}{d}\)

Ta có: \(\frac{a}{b}\)<\(\frac{5a+2c}{5b+2d}\)

\(\Leftrightarrow\)a(5b+2d)<b(5a+2c)

\(\Leftrightarrow\)5ab+2ad<5ab+2bc

\(\Leftrightarrow\)2ad<2bc\(\Leftrightarrow\)ad<bc\(\Leftrightarrow\)\(\frac{a}{b}\)<\(\frac{c}{d}\)(đúng theo giả thiết)

Do vậy:\(\frac{a}{b}\)<\(\frac{5a+2c}{5b+2d}\)

Với lập luận tương tự ta cũng chứng minh được \(\frac{5a+2c}{5b+2d}\)<\(\frac{c}{d}\)

Vậy \(\frac{a}{b}\)<\(\frac{5a+2c}{5b+2d}\)<\(\frac{c}{d}\)

a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}\) = \(\dfrac{3a}{3c}=\dfrac{2b}{2d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}\) (1)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{2b}{2d}=\dfrac{3a-2b}{2c-2d}\) (2)

Từ (1) và(2) ta có:

\(\dfrac{2a+5b}{2c+5d}\) =  \(\dfrac{3a-2b}{3c-2d}\)(đpcm)

a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}\)  ⇒ \(\dfrac{a.b}{c.d}\) = \(\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a.b}{c.d}=\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\) = \(\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)

đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a=bk, c=dk  =>\(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)(1)

=> \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{2k+5}{3k-4}\) ( 2)

từ (1)( 2)=> \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

câu b c/m tg tự 

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{2a+5b}{2c+5d}=\dfrac{2bk+5b}{2dk+5d}=\dfrac{b}{d}\)

\(\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5k}=\dfrac{b}{d}\)

Do đó: \(\dfrac{2a+5b}{2c+5d}=\dfrac{2a-5b}{2c-5d}\)

b: \(\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{b^2k^2-b^2}{b^2k^2+b^2}=\dfrac{k^2-1}{k^2+1}\)

\(\dfrac{c^2-d^2}{c^2+d^2}=\dfrac{d^2k^2-d^2}{d^2k^2+d^2}=\dfrac{k^2-1}{k^2+1}\)

Do đó: \(\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{c^2-d^2}{c^2+d^2}\)