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Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a = b.k; c = d.k
\(\frac{2005a-2006b}{2006c+2007d}=\frac{2005b.k-2006b}{2006d.k+2007.d}=\frac{b\left(2005k-2006\right)}{d\left(2006k+2007\right)}=\frac{b}{d}.\frac{2005k-2006}{2006k+2007}\) (1)
\(\frac{2005c-2006d}{2006a+2007b}=\frac{2005d.k-2006d}{2006b.k+2007b}=\frac{d\left(2005k-2006\right)}{b\left(2006k+2007\right)}=\frac{d}{b}.\frac{2005k-2006}{2006k+2007}\) (2)
Từ (1)(2) => vế trái khác vế phải : Đề sai
Đặt \(\frac{a}{b}=\frac{c}{d}=k\left(1\right)\Rightarrow a=bk;c=dk\)
Thay a và c vào tỉ số \(\frac{2005a-2006b}{2006c+2007d}=\frac{2005c-2006d}{2006a+2007b}\), ta có :
\(\frac{2005a-2006b}{2006c+2007d}=\frac{2005bk-2006b}{2006dk-2007d}=\frac{b\left(2005k-2006\right)}{d\left(2006k+2007\right)}\)
\(\frac{2005c-2006d}{2006a+2007b}=\frac{2005dk-2006d}{2006bk+2007b}=\frac{d\left(2005k-2006\right)}{b\left(2006k+2007\right)}\)
Mà \(\frac{b}{d}\ne\frac{d}{b}\left(b,d\in Z;b\ne d;b,d\ne0\right)\)
=> Sai đề
Ta có: \(\frac{a}{b}=\frac{c}{d}\)=> \(\frac{a}{c}=\frac{b}{d}=\frac{2005a}{2005c}=\frac{2006b}{2006d}=\frac{2006a}{2006c}=\frac{2007b}{2007d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2005a}{2005c}=\frac{2006b}{2006d}=\frac{2006a}{2006c}=\frac{2007b}{2007d}=\frac{2005a-2006b}{2005c-2006d}=\frac{2006a+2007b}{2006c+2007d}\)
=> \(\frac{2005a-2006b}{2006c+2007d}=\frac{2005c-2006d}{2006a+2007b}\)
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)
a) Thay (1) vào đề:
\(VT=\dfrac{a+2006b}{a-2006b}=\dfrac{bk+2006b}{bk-2006b}=\dfrac{b\left(k+2006\right)}{b\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)
\(VP=\dfrac{c+2006d}{c-2006d}=\dfrac{dk+2006d}{dk-2006d}=\dfrac{d\left(k+2006\right)}{d\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)
\(\Rightarrow VT=VP\Leftrightarrow\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}.\)
b) Thay (1) vào đề:
\(VT=\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{2006\left(bk+dk\right)}{2006bk}=\dfrac{bk+dk}{bk}=\dfrac{k\left(b+d\right)}{bk}=\dfrac{b+d}{b}\)
\(VP=\dfrac{b+d}{b}\)
\(\Rightarrow VT=VP\Leftrightarrow\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{b+d}{b}\rightarrowđpcm\).
\(\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}\)
\(\Leftrightarrow\)(a+2006b)(c-2006d)=(c+2006d)(a-2006b)
a(c-2006d)+2006b(c-2006d)=c(a-2006b)+2006d(a-2006b)
ac-2006ad+2006bc-4024036bd=ac-2006bc+2006ad-4024036bd
(ac-2006ad+2006bc-402436bd)-(ac-2006bc+2006ad-4024036bd=0
Suy ra 2 đẳng thức trên =nhau
a.Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
=>\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)(đpcm)
b.Vì\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{a}{c}-1=\dfrac{b}{d}-1\)
=>\(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)(đpcm)
a)\(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
=>\(\dfrac{a}{b}\) -1= \(\dfrac{c}{d}\) -1
=> \(\dfrac{a}{b}\) - \(\dfrac{b}{b}\) = \(\dfrac{c}{d}\) - \(\dfrac{d}{d}\)
=> \(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
*a/b=c/d=k=>a=bk;c=dk
Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3
Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3
=>2a+3b/2a-3b=2c+3d/2c-3d
*a/b=c/d=>a/c=b/d=k
=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)
k^2=a/c.b/d=ab/cd (2)
Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2
*a/b=c/d=>a/c=b/d=k=a+b/c+d
=>k^2=(a+b/c+d)^2
k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2
=>(a+b/c+d)^2=a^2+b^2/c^2+d^2
Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)
a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)
Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(2)
Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Từ \(\dfrac{2005a-2006b}{2006c+2007d}=\dfrac{2005c-2006d}{2006a+2007b}\)
=> \(\dfrac{2005a-2006b}{2005c-2006d}=\dfrac{2006c+2007d}{2006a+2007b}\) (1)
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{d}{b}\)
=> \(\dfrac{2005a}{2005c}=\dfrac{2006b}{2006d}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{2005a}{2005c}=\dfrac{2006b}{2006d}=\dfrac{2005a+2006b}{2005c+2006d}\) (2)
Từ \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{2006a}{2006c}=\dfrac{2007d}{2007b}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{2006a}{2006c}=\dfrac{2007b}{2007d}=\dfrac{2006a-2007d}{2006c-2007b}\) (3)
Từ (1),(2),(3) => \(\dfrac{2005a-2006b}{2006c+2007d}=\dfrac{2005c-2006d}{2006a+2007b}\)