1) \(\frac{x-y}{x+y}=\frac{z-x}{z+x}\)

\(\Leftrightarrow\left(x-y\right)\left(z+x\right)=\left(z-x\right)\left(x+y\right)\)

\(\Leftrightarrow z\left(x-y\right)+x\left(x-y\right)=x\left(z-x\right)+y\left(z-x\right)\)

\(\Leftrightarrow xz-zy+x^2-xy=xz-x^2+yz-xy\)

\(\Leftrightarrow-zy+x^2=-x^2+yz\)

\(\Leftrightarrow-2x^2=-2zy\)

\(\Leftrightarrow x^2=yz\)(đpcm)

b/ VT = (7a – 3b)2 – 4c2 = 49a2- 42ab + 9b2 – 4c2

mà 10a2 = 10b2 + c2 nên c2 = 10a2 – 10b2

nên VT = 49a2 – 42ab + 9b2 – 4(10a2 – 10b2)

= 49a2 – 42ab + 9b2 – 40a2 + 40b2

= 9ª2 – 42ab + 49b2 = (3a – 7b)2 = VP 

Đặt a/b=c/d=k

=>a=bk; c=dk

1: \(\dfrac{2a+15b}{5a-7b}=\dfrac{2\cdot bk+15b}{5\cdot bk-7b}=\dfrac{2k+15}{5k-7}\)

\(\dfrac{2c+15d}{5c-7d}=\dfrac{2dk+15d}{5dk-7d}=\dfrac{2k+15}{5k-7}\)

Do đó: \(\dfrac{2a+15b}{5a-7b}=\dfrac{2c+15d}{5c-7d}\)

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

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

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

hay (a+2c)(b+d)=(a+c)(b+2d)

a: \(\dfrac{2a+15b}{5a-7b}=\dfrac{2c+15d}{5c-7d}\)

\(\Leftrightarrow\left(2a+15b\right)\left(5c-7d\right)=\left(5a-7b\right)\left(2c+15d\right)\)

\(\Leftrightarrow10ac-14ad+75bc-105bd=10ac+75ad-14bc-105bd\)

\(\Leftrightarrow-14ad+75bc=-14bc+75ad\)

=>ad=bc

hay a/b=c/d

b: Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

b) VT = (7a-3b)2 - 4c2 = 49a2 - 42ab + 9b2 - 4c2
mà 10a2 = 10b2 + c2 nên c2 = 10a2 - 10b2
nên VT = 49a2 - 42ab + 9b2 - 4 (10a2 - 10b2)
=49a2 - 42ab + 9b2 - 40a2 + 40b2
=9^d2  - 42ab + 49b2 = (3a - 7b)2 = VT

Ta có: \(\left(x-y\right)\left(x+y\right)=\left(x^2-y^2\right)\)

\(\Rightarrow\left(7a-3b+2c\right)\left(7a-3b-2c\right)=\left(7a-3b\right)^2-\left(2c\right)^2\)

\(=49a^2-42ab+9b^2-4c^2\)

\(=49a^2-42ab+9b^2-4\left(10a^2-10b^2\right)\)

\(=9a^2-2.3.7ab+49b^2=\left(3a-7b\right)^2\left(ĐPCM\right)\)