Cho a, b, c, d khac 0 và \(\frac{a}{b}=\frac{c}{d}\). Hãy chưng minh \(\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\)
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\(\frac{3a+5b}{2a-b}=\frac{3c+5d}{2c-d}\)
<=>\(\left(3a+5b\right)\left(2a-b\right)=\left(3c+5d\right)\left(2c-d\right)\)
<=>\(6ac+10ad-3bc-5bd=6ac+10bc-3ad-5bd\)
<=>\(10ad-3bc=10bc-3ad\)
<=>\(10ad-3bc-10bc+3ad=0\)
<=>\(13ad-13ac=0\)
<=>\(13ad=13ac\)
<=>\(ad=bc\)
<=>\(\frac{a}{b}=\frac{c}{d}\)(đpcm)
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Ta có: \(\frac{3a+5b}{2a-b}=\frac{3c+5d}{2c-d}\)
=> (3a+5b)(2c-d) =(2a-b)(3c+5d)
=> 3a(2c-d) +5b(2c-d) =2a(3c+5d) -b(3c+5d)
=> 6ac -3ad +10bc -5bd =6ac +10ad -3bc -5bd
=>7bc=7ad
=> bc=ad
=> a/b =c/d
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Đặt \(\frac{a}{b}=\frac{c}{d}=v\)
\(\Rightarrow\hept{\begin{cases}a=vb\\c=vd\end{cases}}\)( 1 )
Thay (1) vào vế trái , ta có :
\(VT=\frac{2vb+5b}{3vb-4b}=\frac{b\left(2v+5\right)}{b\left(3v-4\right)}=\frac{2v+5}{3v-4}\)( *)
Thay (1) vào vế phải ta có :
\(VP=\frac{2vd+5d}{3vd-4d}=\frac{2v+5}{3v-4}\)(**)
Từ ( * ) và (** )
=> ĐPCM
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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2a+5d}{3c-4d}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}-\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4d}{3c-4d}\left(=\frac{a}{c}\right)\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Khi đó: \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)
Từ\(\frac{a}{b}\)=\(\frac{c}{d}\)suy ra \(\frac{a}{c}\)=\(\frac{b}{d}\)( t/c TLT)
Áp dụng tính chất của dãy TSBN ta có:\(\frac{a}{c}\)=\(\frac{b}{d}\)=\(\frac{2a+5b}{2c+5d}\)=\(\frac{3a-4b}{3c-4d}\)
Từ \(\frac{2a+5b}{2c+5d}\)=\(\frac{3a-4b}{3c-4d}\) suy ra\(\frac{2a+5b}{3a-4b}\)=\(\frac{2c+5d}{3c-4d}\)(t/c TLT)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a-5b}{2c-5d}\)
mà \(\frac{a}{c}=\frac{3a}{3c}\)
\(\Rightarrow\frac{2a-5b}{2c-5d}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a-5b}{2c-5d}\)
Mà \(\frac{a}{c}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{2c-5d}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\)
(ĐPCM)
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