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Lời giải:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk, c=dk \)
Khi đó:
\(\frac{2002a+2003b}{2002a-2003b}=\frac{2002bk+2003b}{2002bk-2003b}=\frac{b(2002k+2003)}{b(2002k-2003)}=\frac{2002k+2003}{2002k-2003}(1)\)
\(\frac{2002c+2003d}{2002c-2003d}=\frac{2002dk+2002d}{2002dk-2003d}=\frac{d(2002k+2003)}{d(2002k-2003)}=\frac{2002k+2003}{2002k-2003}(2)\)
Từ \((1);(2)\Rightarrow \frac{2002a+2003b}{2002a-2003b}=\frac{2002c+2003d}{2002c-2003d}\)
Ta có đpcm.
Xét tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Gọi giá trị chung của các tỉ số đó là k, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(a=k.b,c=k.d\)
Ta có :
( 1 )
= \(\dfrac{2002a+2003b}{2002a-2003b}=\dfrac{2002kb+2003b}{2002kb-2003b}\)
= \(\dfrac{b.\left(2002k+2003\right)}{b.\left(2002k-2003\right)}=\dfrac{2002k+2003}{2002k-2003}\)
( 2 ) \(\dfrac{2002c+2003d}{2002c-2003d}=\dfrac{2002kd+2003d}{2002kd-2003d}\)
= \(\dfrac{d.\left(2002k+2003\right)}{d.\left(2002k-2003\right)}=\dfrac{2002k+2003}{2002k-2003}\)
Từ ( 1 ) và ( 2 ) => \(\dfrac{2002a+2003b}{2002a-2003b}=\dfrac{2002c+2003d}{2002c-2003d}\)
Ta có : \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2002a}{2002c}=\frac{2003b}{2003d}=\frac{2002a+2003b}{2002c+2003d}=\frac{2002a-2003b}{2002c-2003d}\)
Suy ra : \(\frac{2002a+2003b}{2002a-2003b}=\frac{2002c+2003d}{2002c-2003d}\) (đpcm)
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
Vậy...
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) (1)
Thay (1) vào:
\(\dfrac{a+b}{a-b}=\dfrac{b.k+b}{b.k-b}=\dfrac{b.\left(k+1\right)}{b.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
\(\dfrac{c+d}{c-d}=\dfrac{d.k+d}{d.k-d}=\dfrac{d.\left(k+1\right)}{d.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (3)
Từ (2) và (3) =>\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}=\dfrac{k+1}{k-1}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
=> (a+b)(c-d) = (c+d)(a-b)
=> \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Cách 1 :
Ta có: (a+b)(c-d)=ac-ad+bc-bd (1)
(a-b)(c+d)=ac+ad-bc-bd (2)
Từ giả thiết: \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> ad=bc (3)
Từ (1),(2),(3)=> (a+b)(c-d)=(a-b)(c+d)=>\(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
C2:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k=>a=bk,c=dk\)
Ta có: \(\dfrac{a+b}{a-b}=\dfrac{kb+b}{kb-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\) (1)
\(\dfrac{c+d}{c-d}=\dfrac{kd+d}{kd-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
Từ (1) và (2) =>: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
C3:
Từ giả thiết: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=>\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\Rightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\\ ac+bc-2ad-2bd=ac+ad-2bc-2bd\\ bc-2ad=ad-2bc\\ 3bc=3ad\\ bc=ad\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\)
\(\Leftrightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\)
\(\Leftrightarrow ac-2ad+bc-2bd=ac-2bc+ad-2bd\)
\(\Leftrightarrow2ad+ad=2bc+bc\)
\(\Leftrightarrow3ad=3bc\)
\(\Leftrightarrow ad=bc\rightarrowđpcm\)
AD t/c DTS = nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a+c}{b+d}\left(dpcm\right)\)
*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 \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
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
=> \(\dfrac{a}{b}\) - \(\dfrac{b}{b}\) = \(\dfrac{c}{d}\) - \(\dfrac{d}{d}\)
=> \(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k;c=d.k\)
Xét: \(\frac{2002a+2003b}{2002a-2003b}=\frac{2002bk+2003b}{2002bk-2003b}\)=\(\frac{k+b}{k-b}\) (1)
Mặt khác: \(\frac{2002c+2003d}{2002c-2003d}=\frac{2002dk+2003d}{2002dk-2003d}=\frac{k+d}{k-d}\) (2)
Từ (1) và (2)=> \(\frac{2002a+2003b}{2002a-2003b}=\frac{2002c+2003d}{2002c-2003d}\) (đpcm)