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từ tỉ lệ thức đã cho
=>(3a+4b)(5c-6d)=(3c+4d)(5a-6b)
=>15ac-18ad+20bc-24bd=15ac+20ad-18bc-24bd
=>-18ad+20bc=20ad-18bc
=>-18ad-20ad=-18bc-20bc
=>-38ad=-38bc
=>ad=bc
=>a/b=c/d
=>
\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
=> \(\dfrac{3a+4b}{3c+4d}=\dfrac{5a-6b}{5c-6d}\)
ta có
\(\dfrac{3a+4b}{3c+4d}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a}{b}=\dfrac{c}{d}\)(đpcm)
Ta có:
\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
\(\Leftrightarrow\left(3a+4b\right)\left(5c-6d\right)=\left(3c+4d\right)\left(5a-6b\right)\)
\(\Rightarrow15ac-18ad+20bc-24bd=15ac-18bc+20ad-24bd\)
\(\Rightarrow15ac-15ac-18ad-20ad=-24bd+24bd-18bc-20bc\)
\(\Rightarrow-38ad=-38bc\)
\(\Rightarrow ad=bc\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{bk+b}{b}=\frac{dk+d}{d}\)
Xét VT \(\frac{bk+b}{b}=\frac{b\left(k+1\right)}{b}=k+1\left(1\right)\)
Xét VP \(\frac{dk+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\left(2\right)\)
Từ (1) và (2) -->Đpcm
b)Đặt tương tự ta có:
\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\Leftrightarrow\frac{5bk+3b}{5bk-3b}=\frac{5dk+3d}{5dk-3d}\)
Xét VT \(\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-2}\left(1\right)\)
Xét VP \(\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)
Từ (1) và (2) -->Đpcm
Bạn xem lại đề nhé :)
1) Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
2) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{5}{3}.\frac{a}{b}=\frac{5}{3}.\frac{c}{d}\Rightarrow\frac{5a}{3b}-1=\frac{5c}{3d}-1\Rightarrow\frac{5a-3b}{3b}=\frac{5c-3d}{3d}\)
\(\Rightarrow\frac{3b}{5a-3b}=\frac{3d}{5c-3d}\Rightarrow\frac{6b}{5a-3b}=\frac{6d}{5c-3d}\Rightarrow\frac{6b}{5a-3b}+1=\frac{6d}{5c-3d}+1\)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
1) Vì a/b = c/d
=> a/b + 1 = c/d + 1
=> a + b/b = c + d/d (đpcm)
2) Vì a/b = c/d
=> a/c = b/d
=> 5a/5c = 3b/3d = 5a + 3b/5c + 3d = 5a - 3b/5c - 3d ( theo tc DTSBN )
=> 5a + 3b/5a - 3b = 5c + 3d/5c - 3d
1,a/b=c/d
=>\(\frac{a}{b}+1=\frac{c}{d}+1\)
=>\(\frac{a+b}{b}=\frac{c+d}{d}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{5a+4b}{5a-4b}=\frac{5\cdot bk+4b}{5\cdot bk-4b}=\frac{b\left(5k+4\right)}{b\left(5k-4\right)}=\frac{5k+4}{5k-4}\)
\(\frac{5c+4d}{5c-4d}=\frac{5\cdot dk+4d}{5\cdot dk-4d}=\frac{d\left(5k+4\right)}{d\left(5k-4\right)}=\frac{5k+4}{5k-4}\)
Do đó: \(\frac{5a+4b}{5a-4b}=\frac{5c+4d}{5c-4d}\)
Bài 2:
a: Đặt \(\frac{a}{b}=\frac{b}{c}=k\)
=>b=ck; a=bk=ck^2
\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(ck^2\right)^2+\left(ck\right)^2}{\left(ck\right)^2+c^2}=\frac{c^2k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=k^2\)
\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)
Do đó: \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)
b: Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)
\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)
\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)
Do đó: \(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\)