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\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
\(\dfrac{a}{c}\) = \(\dfrac{b}{d}\)
\(\dfrac{a}{c}\) = \(\dfrac{5a}{5c}\) = \(\dfrac{3b}{3d}\) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}\) = \(\dfrac{5a+3b}{5c+3d}\) (1)
\(\dfrac{a}{c}\) = \(\dfrac{5a-3b}{5c-3d}\) (2)
Kết hợp (1) và (2) ta có:
\(\dfrac{5a+3b}{5c+3d}\) = \(\dfrac{5a-3b}{5c-3d}\)
⇒ \(\dfrac{5a+3b}{5a-3b}\) = \(\dfrac{5c+3d}{5c-3d}\) (đpcm)
b; \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{3a}{3b}\) = \(\dfrac{2c}{2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}\) = \(\dfrac{3a+2c}{3b+2d}\) (đpcm)
a) Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\\\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\end{matrix}\right.\Rightarrowđpcm\)
b) \(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\)\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)
\(\Rightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\) nên \(x-2018=0\Leftrightarrow x=2018\)
a)Đặt \(\dfrac{a}{b}=\dfrac{c}{b}=k\left(k\ne0\right)\)
=> a=bk; c=dk
+) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(1\right)\)
+) \(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) cũng đặt và cm tương tự
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)\(\left(2\right)\)
\(VP=\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
a, Ta có: \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)\(\Rightarrow\dfrac{\left(bk-b\right)^2}{\left(ck-c\right)^2}=\dfrac{bk.b}{dk.d}\)
\(\Rightarrow\dfrac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\)
Vậy \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)
b, Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}\)
\(\Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a+3b}{5c+3d}\)
Lời giải:
Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk$.
Khi đó:
$\frac{5a+3b}{5a-3b}=\frac{5bk+3bk}{5bk-3bk}=\frac{8bk}{2bk}=4(1)$
$\frac{5c+3d}{5c-3d}=\frac{5dk+3dk}{5dk-3dk}=\frac{8dk}{2dk}=4(2)$
Từ $(1); (2)$ suy ra điều phải chứng minh.
Mk chỉ làm 1 câu thôi mấy câu sau tương tự theo cách đó nhoa:v
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^4=\dfrac{b^4}{d^4}\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{bk^4+b^4}{dk^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\Rightarrowđpcm\)
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^4}{c^4}\)=\(\dfrac{b^4}{d^4}\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^4\)(2)
Từ (1) và (2)suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^4\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(đpcm)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a+3b}{5c+3d}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5b}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a-3b}{5c-3d}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)=\(\dfrac{5a+3b}{5a-3b}\)=\(\dfrac{5c+3d}{5c-3d}\) (đpcm)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Do đó: \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{a}{c}\right)^2\)và \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{b}{d}\right)^2\)
=>\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\) và \(\dfrac{ab}{cd}\)=\(\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{7a^2}{7c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{3ab}{3cd}\)=\(\dfrac{7a^2+3ab}{7c^2+3cd}\)(1)
Ta có: \(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=> \(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{7a^2+3ab}{7c^2+3cd}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)=\(\dfrac{7a^2+3ab}{11a^2-8b^2}\)=\(\dfrac{7c^2+3cd}{11c^2-8d^2}\)
đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) thay \(a=bk;c=dk\) ta có
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)
từ (1);(2)\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) thay \(a=bk;c=dk\) ta có
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7(bk)^2+3bkb}{11(bk)^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}\)
\(=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(3)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3dkd}{11\left(dk\right)^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}\)
\(=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(4)
từ (3);(4)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
a) Ta có:
\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}\)
\(\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
b)
\(\frac{2a-b}{2a+b}=\frac{2bk-b}{2bk+b}=\frac{b(2k-1)}{bb(2k+1)}=\frac{2k-1}{2k+1}\)
\(\frac{2c-d}{2c+d}=\frac{2dk-d}{2dk+d}=\frac{d(2k-1)}{d(2k+1)}=\frac{2k-1}{2k+1}\)
\(\Rightarrow \frac{2a-b}{2a+b}=\frac{2c-d}{2c+d}\) (đpcm)
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(1\right)\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)