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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{2a+3c}{3a+4c}=\dfrac{2bk+3dk}{3bk+4dk}=\dfrac{2b+3d}{3b+4d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b, đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
ta có :
\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3b}{5dk-3b}=\frac{d\left(5k+3b\right)}{d\left(5k-3b\right)}=\frac{5k+3b}{5k-3b}\)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3b}\)
xin lỗi nha viết gấp quá quên cả kết luận :))
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Ta có:
\(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{3a}{3b}=\frac{3c}{3d}\)=>\(\frac{3a}{3c}=\frac{3b}{3d}\) ; \(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{4a}{4b}=\frac{4c}{4d}\)=>\(\frac{4a}{4c}=\frac{4b}{4d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{3a}{3c}=\frac{3b}{3d}=\frac{3a+3b}{3c+3d}\) ; \(\frac{4a}{4c}=\frac{4b}{4d}=\frac{4a+4b}{4c+4d}\)
Mà \(\frac{3a}{3b}=\frac{3b}{3d}=\frac{4a}{4c}=\frac{4b}{4d}\)
=>\(\frac{3a+3b}{3c+3d}=\frac{4a+4b}{4c+4d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{3a+4b}{3c+4d}\left(1\right)\)
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{6a}{6c}=\dfrac{7b}{7d}=\dfrac{6a+7b}{6c+7d}\left(2\right)\)
\(\left(1\right)\left(2\right)\RightarrowĐpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
t.i.c.k mik mik t.i.c.k lại