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\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
=> (3a + 5b)(3c - 5d) = (3a - 5b)(3c + 5d)
=> 9ac - 15ad + 15bc - 25bd = 9ac + 15ad - 15bc - 25bd
=> 9ac - 15ad + 15bc - 25bd - (9ac + 15ad - 15bc - 25bd) = 0
=> 9ac - 15ad + 15bc - 25bd - 9ac - 15ad + 15bc + 25bd = 0
=> (9ac - 9ac) + (-15ad - 15ad) + (15bc + 15bc) + (-25bd + 25bd) = 0
=> -30ad + 30bc = 0
=> -30ad = -30bc
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)
Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)
c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)
Theo dãy tỉ số (=) ta* có:
\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3a+3b+3c+3d}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)
=> a = 1/3 . 3b = b (1)
=> b = 1/3 . 3c = c (2)
=> c = 1/3 . 3d = d (3)
Từ(1) (2) và (3) =. a = b= c =d => ĐPCM
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: \(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)
\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{c\left(2c-a\right)}{d\left(2d-b\right)}=\dfrac{dk}{d}\cdot\dfrac{2dk-bk}{2d-b}=k^2\)
=>\(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)
\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3a+3b+3c+3d}=\frac{1}{3}.\) (T/c dãy tỷ số bằng nhau)
=> \(\frac{a}{3b}=\frac{1}{3}\Rightarrow\frac{a}{b}=1\Rightarrow a=b\)
Làm tương tự sẽ rút ra a=b=c=d