Cho a/b=c/d, cho mình rằng 7a+2c/7b+2d = 7a-2c/7b-2d
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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; \(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)


Ta có: \(\left(x-y\right)\left(x+y\right)=\left(x^2-y^2\right)\)
\(\Rightarrow\left(7a-3b+2c\right)\left(7a-3b-2c\right)=\left(7a-3b\right)^2-\left(2c\right)^2\)
\(=49a^2-42ab+9b^2-4c^2\)
\(=49a^2-42ab+9b^2-4\left(10a^2-10b^2\right)\)
\(=9a^2-2.3.7ab+49b^2=\left(3a-7b\right)^2\left(ĐPCM\right)\)
\(\frac{a}{b}=\frac{c}{d}=\frac{7a}{7b}=\frac{2c}{2d}=\frac{7a+2c}{7b+2d}\left(1\right)\) (Tính chất dãy tỷ số bằng nhau)
\(\frac{a}{b}=\frac{c}{d}=\frac{7a}{7b}=\frac{2c}{2d}=\frac{7a-2c}{7b-2d}\left(2\right)\) (lý do như trên)
Từ (1) và (2) \(\Rightarrow\frac{7a+2c}{7b+2d}=\frac{7a-2c}{7b-2d}\)