giả sử a/2002 = b/2003 = c/2004 = k
=> a = 2002k ; b=2003k và c=2004k
=> 4(a-b)(b-c) = 4(2002k - 2003k)(2003k - 2004k)
=> 2(a-b)(b-c) = 4k^2 (1)
Ta có (c-a)^2 = (2004k - 2002k)^2 = 4k^2 (2)
từ (1) và (2) ta có 2(a-b)(b-c) = (c-a)^2

Bài 1:

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)

a: Đặt a/b=b/c=c/d=k

=>a=bk; b=ck; c=dk

=>a=bk; b=dk^2; c=dk

=>a=dk^3; b=dk^2; c=dk

\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=k^3\)

\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)

=>\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)

c: Đặt a/2003=b/2004=c/2005=k

=>a=2003k; b=2004k; c=2005k

4(a-b)(b-c)=(c-a)^2

=>4(2004k-2003k)(2005k-2004k)=(2005k-2003k)^2

=>4*k*k=(2k)^2(luôn đúng)

=>ĐPCM

5a

Ta có \(\dfrac{a}{b}=\dfrac{a^2}{b^2}\) ; \(\dfrac{c}{d}=\dfrac{c^2}{d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\)=> \(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=>\(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=\(\dfrac{a^2+c^2}{b^2+d^2}\)(T/c cuả dãy tỉ số bằng nhau)

=> ĐPCM

Bài 1:

Ta có: \(\dfrac{U+2}{U-2}=\dfrac{V+3}{V-3}\)

\(\Rightarrow\left(U+2\right)\left(V-3\right)=\left(U-2\right)\left(V+3\right)\)

\(\Rightarrow\left(U+2\right)V-\left(U+2\right)3=\left(U-2\right)V+\left(U-2\right)3\)

\(\Rightarrow UV+2V-3U-6=UV-2V+3U-6\)

\(\Rightarrow2V-3U=3U-2V\)

\(\Rightarrow2U+2U=3V+3V\)

\(\Rightarrow2\times2U=2\times3V\)

\(\Rightarrow2U=3V\)

\(\Rightarrow\dfrac{U}{3}=\dfrac{V}{2}\)

(Như mình đã nói ở Câu hỏi của ARMY, cách này tuy dài nhg chặt hơn)

(Tiếp tục nè)

\(\Rightarrow\dfrac{U^2}{9}=\dfrac{V^2}{4}=\dfrac{U^2+V^2}{9+4}=\dfrac{52}{13}=4\)

\(\Rightarrow U^2=4\times9=36\Rightarrow U=\pm6\)

\(V^2=4\times4=16\Rightarrow V=\pm4\)

Vì \(\dfrac{U}{3}=\dfrac{V}{2}\Rightarrow U,V\) cùng dấu

\(\Rightarrow\left(U,V\right)=\left(6;4\right)\) và \(\left(U,V\right)=\left(-6;-4\right)\)

Vậy...

Bài 2:

Vì:\(\dfrac{x}{y}=\dfrac{z}{t}\Rightarrow\dfrac{x}{z}=\dfrac{y}{t}=\dfrac{x+y}{z+t}\Rightarrow\left(\dfrac{x}{z}\right)^2=\left(\dfrac{y}{t}\right)^2=\left(\dfrac{x+y}{z+t}\right)^2\left(1\right)\)

Ta còn có: \(\dfrac{x}{z}=\dfrac{y}{t}\Rightarrow\left(\dfrac{x}{z}\right)^2=\left(\dfrac{y}{t}\right)^2=\dfrac{xy}{zt}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow\).....

Chúc bn học tốt

P/S: Những chỗ mình ghi chấm chấm có nghĩa là bn tự kết nốt

4.a

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)

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

b: Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

a: a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)

b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)

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

c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)

\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)

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

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)