Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Sửa: \(\dfrac{3a^2+10b^2-ab}{7a^2+b^2+5ab}=\dfrac{3b^2k^2+10b^2-b^2k}{7b^2k^2+b^2+5b^2k}=\dfrac{b^2\left(3k^2+10-k\right)}{b^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(1\right)\)
\(\dfrac{3c^2+10d^2-cd}{7c^2+d^2+5cd}=\dfrac{3d^2k^2+10d^2-d^2k}{7d^2k^2+d^2+5d^2k}=\dfrac{d^2\left(3k^2+10-k\right)}{d^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(2\right)\)
\(\left(1\right)\left(2\right)\RightarrowĐpcm\)
Bạn tham khảo tại link sau:
Câu hỏi của Nguyễn Thanh Huyền - Toán lớp 7 | Học trực tuyến
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)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
Hình như sai sai
a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}\) = \(\dfrac{3a}{3c}=\dfrac{2b}{2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}\) (1)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{2b}{2d}=\dfrac{3a-2b}{2c-2d}\) (2)
Từ (1) và(2) ta có:
\(\dfrac{2a+5b}{2c+5d}\) = \(\dfrac{3a-2b}{3c-2d}\)(đpcm)
a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}\) ⇒ \(\dfrac{a.b}{c.d}\) = \(\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a.b}{c.d}=\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\) = \(\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)
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