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![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3b+3c+3d+3x}=\dfrac{a+b+c+d}{3.\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow a=\dfrac{1}{3}.3b=b\\ \Rightarrow b=\dfrac{1}{3}.3c=c\\ \Rightarrow c=\dfrac{1}{3}.3d=d\\ \Rightarrow d=\dfrac{1}{3}.3a=a\)
➩\(\text{a=b=c=d}\)
Tick cho mình nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo tính chất của dãy tỉ số bằng nhau :
\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Vì a + b + c + d khác 0 . Ta có :
\(a=\dfrac{1}{3}.3b=b\)(1)
\(b=\dfrac{1}{3}.3c=c\)(2)
\(c=\dfrac{1}{3}.3d=d\)(3)
\(d=\dfrac{1}{3}.3a=a\)(4)
Từ (1);(2);(3) và (4)
=> a = b = c = d
![](https://rs.olm.vn/images/avt/0.png?1311)
d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)
\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)
Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
- Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
+ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
+ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\)
Theo đề ta có: \(a:b=c:d\); \(b,d\ne0,b\ne\pm d\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\end{matrix}\right.\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3\left(b+c+d+a\right)}=\dfrac{1}{3}\)
\(\dfrac{a}{3b}=\dfrac{1}{3}\Rightarrow a=b\) __( 1 )__
\(\dfrac{b}{3c}=\dfrac{1}{3}\Rightarrow b=c\) __( 2 )__
\(\dfrac{c}{3d}=\dfrac{1}{3}\Rightarrow c=d\) __( 3 )__
\(\dfrac{d}{3a}=\dfrac{1}{3}\Rightarrow d=a\) __ ( 4 )__
Từ ( 1 ), ( 2 ), ( 3 ), ( 4 ) suy ra: \(a=b=c=d\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}=\dfrac{c}{d}=>\)\(ad=bc\)
\(\dfrac{a}{a-b}=\dfrac{ad}{d\left(a-b\right)}=\dfrac{bc}{ad-bd}=\dfrac{bc}{bc-bd}=\dfrac{bc}{b\left(c-d\right)}\dfrac{c}{c-d}\)
![](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)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4bk+3dk}{4b+3d}=k\)
\(\dfrac{4a-3c}{4b-3d}=\dfrac{4bk-3dk}{4b-3d}=k\)
Do đó: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4a-3c}{4b-3d}\)
Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)
Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)
+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)
Thanks.