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Ta có:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a) \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b) \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c) \(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2.\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2\right)+1}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
c) có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a^2}{^{c^2}}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1) và (2) có \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
các câu còn lại bạn tự làm đi! HI.......
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\) thì \(a=bk,c=dk\).
\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\\ \frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)
Do đó: \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Xem ở lick này nhé (mình gửi cho)
Học tốt!!!!!!!!!!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{2b}{2d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{b}{d}=\frac{2a}{2c}=\frac{2a+b}{2c+d}=\frac{a}{c}=\frac{2b}{2d}=\frac{a-2b}{c-2d}\)
\(\Rightarrow\frac{2a+b}{2c+d}=\frac{a-2b}{c-2d}\)
\(\Rightarrow\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\left(đpcm\right)\)
b ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}=\frac{a-c}{b-d}\)
\(\Rightarrow\frac{a+2c}{b+2d}=\frac{a-c}{b-d}\)
\(\Rightarrow\left(a+2c\right)\left(b-d\right)=\left(a-c\right)\left(b+2d\right)\left(đpcm\right)\)
Chúc bạn học tốt !!!
\(\frac{a}{c}=\frac{b}{d}\)
suy ra\(\frac{2a}{2c}=\frac{b}{d}=\frac{2a+b}{2c+d}\left(1\right)\)
\(\frac{a}{c}=\frac{2b}{2d}=\frac{a-2b}{c-2d}\left(2\right)\)
\(tu\left(1\right)\left(2\right)suyra\)\(\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
ta có : \(\frac{4a-3b}{a}=\frac{4bk-3b}{bk}=\frac{b\left(4k-3\right)}{bk}=\frac{4k-3}{k}\)
\(\frac{4c-3d}{c}=\frac{4dk-3d}{dk}=\frac{d\left(4k-3\right)}{dk}=\frac{4k-3}{k}\)
\(\Rightarrow\frac{4a-3b}{a}=\frac{4c-3d}{c}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{ab}{cd}=\frac{bk.b}{dk.b}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)
từ\(\left(1\right)\)và\(\left(2\right)\)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{2c^2}{2d^2}=\frac{ac}{bd}\)
Ta có : \(\frac{2c^2}{2d^2}=\frac{ac}{bd}=\frac{2c^2-ac}{2d^2-bd}\)
Vậy \(\frac{a^2}{b^2}=\frac{2c^2-ac}{2d^2-bd}\)