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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)
Bạn tham khảo tại link sau:
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
Gọi a/b=c/d=k =>a=bk;c=dk
=>\(\frac{7a^2+5ac}{7a^2-5ac}=\frac{7\left(bk\right)^2+5\left(bk\right)\left(dk\right)}{7\left(bk\right)^2-5\left(bk\right)\left(dk\right)}=\frac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\frac{k^2\left(7b^2+5bd\right)}{k^2\left(7b^2-5bd\right)}=\frac{7b^2+5bd}{7b^2-5bd}\)
Vậy \(\frac{7a^2+5ac}{7a^2-5ac}=\frac{7b^2+5bd}{7b^2-5bd}\)
Đỗ Lê Tú Linh, cảm ơn bạn nhiều, mình cũng làm như thế nhưng lại quên không thay c=dk. Giờ mình biết làm rồi
Gọi a/b=c/d=k => a=bk; c=dk
=> \(\frac{7a^2+5ac}{7a^2-5ac}\) = \(\frac{7\left(bk\right)^2+5\left(bk\right)\left(dk\right)}{7\left(bk\right)^2-5\left(bk\right)\left(dk\right)}\) = \(\frac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}\) = \(\frac{k^2\left(7b^2+5bd\right)}{k^2\left(7b^2-5bd\right)}\) = \(\frac{7b^2+5bd}{7b^2-5bd}\)
Vậy \(\frac{7a^2+5ac}{7a^2-5ac}\) = \(\frac{7b^2+5bd}{7b^2-5bd}\)
\(\frac{a}{b}=\frac{c}{d}\)
=> \(\frac{a}{c}=\frac{b}{d}\)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{ab}{cd}=\frac{5ab}{5cd}=\frac{a^2-b^2}{c^2-d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{3a^2}{3c^2}=\frac{5ab}{5cd}=\frac{a^2-b^2}{c^2-d^2}=\frac{3a^2+5ab}{3c^2+5cd}\)
=> \(\frac{a^2-b^2}{c^2-d^2}=\frac{3a^2+5ab}{3c^2+5cd}\)
=> \(\frac{3a^2+5ab}{a^2-b^2}=\frac{3c^2+5cd}{c^2-d^2}\)
=> Đpcm
đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk
c=dk
bạn thay vào rùi làm tiếp
1/
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b,\(\frac{a}{b}=\frac{c}{d}=\frac{4a}{4b}=\frac{7c}{7d}=\frac{4a+7c}{4b+7d}\)
2/
Gọi số học sinh tham gia của mỗi lớp lần lượt là a,b,c
Ta có: \(2a=3b=4c\)
\(\Rightarrow\frac{2a}{12}=\frac{3b}{12}=\frac{4c}{12}\Rightarrow\frac{a}{6}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{6+4+3}=\frac{130}{13}=10\)
=> a/6 = 10 => a = 60
b/4 = 10 => b = 40
c/3 = 10 => c = 30
Vậy số học sinh mỗi lớp lần lượt là 60 hs, 40 hs, 30hs
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
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