Question

Cho tỉ lệ thức : \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\). CMR ta có:
\(\dfrac{\text{2002a+2003b}}{\text{2002a - 2003b}}\) = \(\dfrac{\text{2002c+2003d}}{\text{2002c−2003d}}\)

Answer 1

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.

Answer 2

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 ) => \(\dfrac{2002a+2003b}{2002a-2003b}=\dfrac{2002c+2003d}{2002c-2003d}\)

Answer 3

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2002a}{2002c}=\dfrac{2003b}{2003d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{2002a}{2002c}=\dfrac{2003b}{2003d}=\dfrac{2002a+2003b}{2002c+2003d}=\dfrac{2002a-2003b}{2002c-2003d}\)

Từ \(\dfrac{2002a+2003b}{2002c+2003d}=\dfrac{2002a-2003b}{2002c-2003d}\) =>\(\dfrac{2002a+2003b}{2002a-2003b}=\dfrac{2002c+2003d}{2002c-2003d}\)