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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 :
\(\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Sửa chút, chỗ mẫu 11c + 3b thành 11c +3d
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{c}=\frac{b}{d}=\frac{11a}{11c}=\frac{3b}{3d}=\frac{11a+3b}{11c+3d}\\\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{11b}{11d}=\frac{3a-11b}{3c-11d}\end{cases}}\)
\(\Rightarrow\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)
Vậy \(\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)
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Câu hỏi của Nguyễn Thanh Huyền - Toán lớp 7 | Học trực tuyến
a) \(\frac{4a-3b}{a}=\frac{4c-3d}{c}\Leftrightarrow4-\frac{3b}{a}=4-\frac{3d}{c}\)
\(\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
b) \(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\Leftrightarrow\frac{9\left(9999-11d\right)-88880c}{9999c-11d}=\frac{9\left(9999a-11b\right)-88880a}{9999a-11b}\)
\(\Leftrightarrow9+\frac{-88880c}{9999c-11d}=9+\frac{-88880a}{9999a-11b}\)
\(\Leftrightarrow\frac{c}{9999c-11d}=\frac{a}{9999a-11b}\)
\(\Leftrightarrow\frac{9999c-11d}{c}=\frac{9999a-11b}{a}\)
\(\Leftrightarrow9999-\frac{11d}{c}=9999-\frac{11b}{a}\Leftrightarrow\frac{d}{c}=\frac{b}{a}\Leftrightarrow\frac{c}{d}=\frac{a}{b}\)
câu b hình như đề sai
\(\dfrac {1111c-99d}{9999c-11d}=\dfrac {1111a-99b}{9999a-11b}\)
Sửa đề:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7bk-11b}{4bk+5b}=\dfrac{7k-11}{4k+5}\)
\(\dfrac{7c-11d}{4c+5d}=\dfrac{7dk-11dk}{4dk+5d}=\dfrac{7k-11}{4k+5}\)
Do đó: \(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
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:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\left(đpcm\right)\)
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Từ \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) \(\Rightarrow\dfrac{c-d}{c+d}=\dfrac{a-b}{a+b}\)
b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=b.k;b=d.k\)
Thay :
(1) : \(\dfrac{3a+2b}{3a-2b}=\dfrac{3bk+2b}{3bk-2b}=\dfrac{b.\left(3.k+2\right)}{b.\left(3.k-2\right)}=\dfrac{3.k+2}{3.k-2}\)
(2) : \(\dfrac{3c+2d}{3c-2d}=\dfrac{3dk+2d}{3dk-2d}=\dfrac{d.\left(3.k+2\right)}{d.\left(3.k-2\right)}=\dfrac{3.k+2}{3.k-2}\)
Do đó : \(\dfrac{3a+2b}{3a-2b}=\dfrac{3c+2d}{3c-2d}\)