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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow a=bk,c=dk\)
Thay a = bk, c = dk vào \(\frac{7a^2+3ab}{2a^2-ab}\)và \(\frac{7c^2+3cd}{2c^2-cd}\), ta có:
\(\frac{7a^2+3ab}{2a^2-ab}=\frac{7\left(bk\right)^2+3.bk.b}{2\left(bk\right)^2-bk.b}=\frac{7b^2k^2+3b^2k}{2b^2k^2-b^2k}=\frac{b^2k\left(7k+3\right)}{b^2k\left(2k-1\right)}=\frac{7k+3}{2k-1}\)
\(\frac{7c^2+3cd}{2c^2-cd}=\frac{7\left(dk\right)^2+3.dk.d}{2\left(dk\right)^2-dk.d}=\frac{7d^2k^2+3d^2k}{2d^2k^2-d^2k}=\frac{d^2k\left(7k+3\right)}{d^2k\left(2k-1\right)}=\frac{7k+3}{2k-1}\)
\(\Rightarrow\frac{7a^2+3ab}{2a^2-ab}=\frac{7c^2+3cd}{2c^2-cd}\left(đpcm\right)\)
Đặt a/b=c/d=k thì a=bk, c=dk
*7a2 +3ab/2a2-ab=7b2k2+3b2k/2b2k2-b2k=b2k(7k+3)/b2k(2k-1)=7k+3/2k-1 (1)
Tương tự 7c2+3cd/2c2-cd=7k+3/2k-1 (2)
từ (1) và (2) suy ra :
7a2+3ab2a2−ab =7c2+3cd2c2−cd
Ta có : \(\frac{a}{b}=\frac{c}{d}\)=> \(\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\)=> \(\hept{\begin{cases}a=ck\\d=dk\end{cases}}\)
Khi đó, ta có : \(\frac{2\left(ck\right)^2-3\left(ck\right)\left(dk\right)+5\left(dk\right)^2}{2\left(dk\right)^2+3\left(ck\right)\left(dk\right)}=\frac{2c^2k^2-3cdk^2+5d^2k^2}{2d^2k^2+3cdk^2}=\frac{\left(2c^2-3cd+5d^2\right)k^2}{\left(2d^2+3cd\right)k^2}\)
= \(\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)(Đpcm)
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đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Thay a và c vào VP và VT sẽ bằng nhau
Ai help me vs
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\)
Ta có:
\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2\left(bk\right)^2-2bkb+5b^2}{2b^2+3bkb}=\dfrac{2b^2k^2-2b^2k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(1\right)\)
\(\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(2\right)\)
Từ (1) và (2) suy ra:
\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2.\left(bk\right)^2-3.bk.b+5.b^2}{2b^2+3.bk.b}\)=\(\frac{2.b^2.k^2-3.k.b^2+5.b^2}{2.b^2+3.b^2.k}=\frac{b^2\left(2.k^2-3.k+5\right)}{b^2\left(2+3.k\right)}=\frac{2.k^2-3.k+5}{2+3.k}\)
\(\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2.\left(dk\right)^2-3.dk.d+5.d^2}{2.d^2+3.dk.d}\)\(=\frac{2.d^2.k^2-3.d^2.k+5.d^2}{2.d^2+3.d.k.d}\)=\(\frac{d^2\left(2.k^2-3.k+5\right)}{d^2\left(2+3.k\right)}=\frac{2.k^2-3.k+5}{2+3.k}\)
=> bằng nhau
Mình hướng dẫn thôi. Chứ giờ đang bận.
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\).Rồi thay a = kb; c=kd vào từng vế. Thấy hai vế bằng nhau => đpcm
\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}=\frac{2a^2-3ab+5b^2}{2b^2-3cd+5d^2}=\frac{2b^2+3ab}{2d^2+3cd}\)
\(=>\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)