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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 \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
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 a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
=>\(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
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 a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
Do đó: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b: \(\dfrac{7a^2+8ab}{11a^2-8b^2}=\dfrac{7b^2k^2+8\cdot bk\cdot b}{11\cdot b^2\cdot k^2-8b^2}=\dfrac{7k^2+8k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+8\cdot dk\cdot d}{11\cdot d^2\cdot k^2-8d^2}=\dfrac{7k^2+8k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2+8ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
1,
Giải:
a, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\) (1)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
\(\Rightarrowđpcm\)
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\)
\(\Rightarrow\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\)
\(\dfrac{k-1}{k}=\dfrac{k-1}{k}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\rightarrowđpcm\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{5a}{3b}=\dfrac{5c}{3d}\)
hay \(\dfrac{5a}{5c}=\dfrac{3b}{3d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Leftrightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
hay \(\dfrac{5a+3n}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)(đpcm)
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)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3b}{3d}=\dfrac{5a}{5c}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\\ \Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
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.