Helpp 

a: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5bk+2b}{5bk-2b}=\dfrac{5k+2}{5k-2}\)

\(\dfrac{5c+2d}{5c-2d}=\dfrac{5dk+2d}{5dk-2d}=\dfrac{5k+2}{5k-2}\)

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

giải nốt câu b

Áp dụng tính chất của dãy tỉ số = nhau ta có:

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

=> a = b = c = d

=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)

D = 1 + 1 + 1 + 1 = 4

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-2=\frac{c}{d}-2\Rightarrow\frac{a-2b}{b}=\frac{c-2d}{d}\)

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 a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+2b}{a}=\dfrac{bk+2b}{bk}=\dfrac{k+2}{k}\)

\(\dfrac{c+2d}{c}=\dfrac{dk+2d}{dk}=\dfrac{k+2}{k}\)

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

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{bk+2b}{dk+2d}=\dfrac{b\left(k+2\right)}{d\left(k+2\right)}=\dfrac{b}{d}\)

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

\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{a-2b}{c-2d}\rightarrowđpcm\)

ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow4ad=4bc\Leftrightarrow2ad+2ad=2bc+2bc\)

\(\Leftrightarrow2ad-2bc=2bc-2ad\Leftrightarrow ac+2ad-2bc-4bd=ac+2bc-2ad-4bd\)

\(\Leftrightarrow\left(c+2d\right)\left(a-2b\right)=\left(a+2b\right)\left(c-2d\right)\Leftrightarrow\dfrac{a+2b}{c+2d}=\dfrac{a-2b}{c-2d}\left(đpcm\right)\)

E=(-a-b+c+d)-(d+c-b-2a)

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

E=-a+(-)b+c+d+(-d)+(-c)+b+2a

E=-a+(-b)+c+d+(-d)+(-c)+b+2a

E=(2a-a)+(-b+b)+(-d+d)+(-c+c)=a+0+0+0=a

thanks nhiều nha ĐỨC THỊNH

F=(a-2b-c-2d)-(3d-2c-3b+a)+15

F=a-2b-c-2d-3d+2c+3b-a+15

F=a+(-2b)+(-c)+(-2d)+(-3d)+2c+3b+(-a)+15

F=(-2b+3b)+(-c+2c)+[-2d+(-3d)]+(-a+a)+15

F=b+c+(-5d)+0+15=b+c+(-5d)+15