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\(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\Rightarrow\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}\) (1)
Nhân tư và mẫu vế trái (1) với 3 và vế phải với 13 ta được:
\(\dfrac{2a+13b}{2c+13d}=\dfrac{14a+91b}{14c+91d}=\dfrac{39a-91b}{39c-91d}\)
=\(\dfrac{\left(14a+91b\right)+\left(39a-91b\right)}{\left(14c+91d\right)+\left(39c-91d\right)}=\dfrac{53a}{53c}=\dfrac{a}{c}\) (2)
Nhân tử và mẫu vế trái (1) với 3 và vế phải với 2 ta được:
\(\dfrac{2a+13b}{2c+13d}=\dfrac{6a+39b}{6c+39d}=\dfrac{6a-14b}{6c-14d}=\dfrac{53b}{53d}=\dfrac{b}{d}\) (3)
Từ (2) và (3) suy ra :
\(\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Nguyễn Huy Tú chắc làm sai rồi
Chứng minh:
Ta có: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)
\(\Rightarrow\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}=\dfrac{2a+13b+3a-7b}{2c+13d+3c-7d}=\dfrac{5a+6b}{5c+6d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left\{{}\begin{matrix}a=b\\c=d\end{matrix}\right.\Rightarrow\dfrac{a}{a}=\dfrac{c}{c}\)
\(\Rightarrow\dfrac{a+a}{a}=\dfrac{c+c}{c}\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Vậy \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\) (Đpcm)
\(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)
\(\Leftrightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)
\(\Leftrightarrow6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)
\(\Leftrightarrow-53ad=-53bc\)
=>ad=bc
hay a/b=c/d
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Suy ra : \(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b.\left(2k+13\right)}{b.\left(3k-7\right)}=\frac{2k+13}{3k-7}\)
\(\frac{2c+13d}{3c-7d}=\frac{2dk+13d}{3dk-7d}=\frac{d\left(2k+13\right)}{d\left(3k-7\right)}=\frac{2k+13}{3k-7}\)
Vậy \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\) Khi : \(\frac{a}{b}=\frac{c}{d}\)
ta có : \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)
<=> (2a+13b)(3c-7d)=(2c+13d)(7a-7b)
<=>6ac-14ad+39bc-91bd=6c-14bc+39ab-91bd
<=>39bc-14ab=39ab-14bc
<=> bc=ab
<=>\(\frac{a}{b}=\frac{c}{d}\)
Ta có: \(\frac{2a+13b}{3a-7c}=\frac{2c+13d}{3a-7d}\)
\(\Rightarrow\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}=\frac{2a+13b+3a-7b}{2c+13d+3c-7d}=\frac{5a+6b}{5c+6d}\)
\(\Rightarrow\frac{5a+6b}{5c+6d}\Rightarrow\frac{5a}{5c}=\frac{6b}{6d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\left(đpcm\right)\)
\(\dfrac{2a+13b}{3a-7b}\)=\(\dfrac{2c+13d}{3c-7d}\)
CMR:\(\dfrac{a}{b}=\dfrac{c}{d}\)
mn giải giúp cốm
Ta có: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)
\(\Leftrightarrow\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}\)
\(\Leftrightarrow\dfrac{a}{c}+\dfrac{b}{d}=\dfrac{a}{c}-\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
hay \(\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)
\(\Rightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)
\(\Rightarrow6ac+39bc-14ad-91bd=6ac+39ad-14bc-91bd\)
\(\Rightarrow6ac-6ac+39bc+14bc-14ad-39ad-91bd+91bd=0\)
\(\Rightarrow53bc-53ad=0\)
\(\Rightarrow53bc=53ad\)
\(\Rightarrow bc=ad\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\rightarrowđpcm.\)
\(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)
\(\Leftrightarrow\)(2a+13b)(3c-7d)=(2c+13d)(3a-7b)
2a(3c-7d)+13b(3c-7d)=2c(3a-7b)+13d(3a-7b)
6ac-14ad+39bc-91bd=6ac-14bc+39ad+91bd
14ad+39bc+91bd=14bc+39ad+91bd
14ad+39bc=14bc+39ad
39bc=14bc+39ad-14ad
39bc=14bc+25ad
39bc-14bc=25ad
25bc=25ad
bc=ad
Ta có: Điều đề bài cho:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\left(đpcm\right)\)