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Câu hỏi của Nguyễn Thanh Huyền - Toán lớp 7 | Học trực tuyến

Ta có:

\(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\frac{3a^2+5ab}{7a^2-10b^2}=\frac{3.\left(bk\right)^2+5.bkb}{7\left(bk\right)^2-10b^2}=\frac{3b^2k^2+5kb^2}{7b^2k^2-10b^2}=\frac{kb^2\left(3k+5\right)}{b^2\left(7k^2-10\right)}=\frac{k\left(3k+5\right)}{\left(7k^2-10\right)}\left(1\right)\)

\(\frac{3c^2+5cd}{7c^2-10d^2}=\frac{3.\left(dk\right)^2+5dkd}{7\left(dk\right)^2-10d^2}=\frac{3d^2k^2+5kd^2}{7d^2k^2-10d^2}=\frac{kd^2\left(3k+5\right)}{d^2\left(7k^2-10\right)}=\frac{k\left(3k+5\right)}{\left(7k^2-10\right)}\left(2\right)\)

Từ (1) và (2)

⇒ĐPCM

Lời giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)

Khi đó :

\(\frac{3a^2+5ab}{7a^2-10b^2}=\frac{3(bt)^2+5.bt.b}{7(bt)^2-10b^2}=\frac{b^2(3t^2+5t)}{b^2(7t^2-10)}=\frac{3t^2+5t}{7t^2-10}\)

\(\frac{3c^2+5cd}{7c^2-10d^2}=\frac{3(dt)^2+5dt.d}{7(dt)^2-10d^2}=\frac{d^2(3t^2+5t)}{d^2(7t^2-10)}=\frac{3t^2+5t}{7t^2-10}\)

\(\Rightarrow \frac{3a^2+5ab}{7a^2-10b^2}=\frac{3c^2+5cd}{7c^2-10d^2}\) (đpcm)

Tính chất dãy tỉ số bằng nhau

a/ Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

\(\dfrac{2a+7b}{3a-4b}=\dfrac{2bk+7b}{3bk-4b}=\dfrac{b\left(2k+7\right)}{b\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\left(1\right)\)

\(\dfrac{2c+7d}{3c-4d}=\dfrac{2dk+7d}{3dk-4d}=\dfrac{d\left(2k+7\right)}{d\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\)\(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ tương tự

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\left(1\right)\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\left(2\right)\)

\(\left(1\right)\left(2\right)\RightarrowĐpcm\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)

\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)

\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)

\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)

\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)

\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)

\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)

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