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:

\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)

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

Ta có: \(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2b^2k^2-3b^2k+5d^2}{2b^2+3b^2k}\)

\(=\dfrac{b^2k\left(2k-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{k\left(2k-3+5\right)}{2+3k}\) (1)

\(\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}\)

\(=\dfrac{d^2k\left(2k-3+5\right)}{d^2\left(2+3k\right)}=\dfrac{k\left(2k-3+5\right)}{2+3k}\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\left(đpcm\right)\)

Đặt :

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

\(\Leftrightarrow a=bk;c=dk\)

\(VP=\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2\left(bk\right)^2-3bkb+5b^2}{2b^2+3bkb}=\dfrac{2b^2.k^2-2b^2.k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(1\right)\)

\(VT=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2\left(dk\right)^2-3dkd+5d^2}{2\left(dk\right)^2+3dkd}=\dfrac{2.d^2.k^2-3d^2.k+5.d^2}{2.d^2.k^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(2\right)\)

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

thôi mk làm đc rùi ko cần nữa nha ^-^

Đặt:

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

\(\Rightarrow\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\)\(\Rightarrow\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\)\(\Rightarrow\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)

\(\Rightarrowđpcm\)

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{2a+5b}{2c+5d}=\dfrac{2bk+5b}{2dk+5d}=\dfrac{b}{d}\)

\(\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5k}=\dfrac{b}{d}\)

Do đó: \(\dfrac{2a+5b}{2c+5d}=\dfrac{2a-5b}{2c-5d}\)

b: \(\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{b^2k^2-b^2}{b^2k^2+b^2}=\dfrac{k^2-1}{k^2+1}\)

\(\dfrac{c^2-d^2}{c^2+d^2}=\dfrac{d^2k^2-d^2}{d^2k^2+d^2}=\dfrac{k^2-1}{k^2+1}\)

Do đó: \(\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{c^2-d^2}{c^2+d^2}\)

Tham khảo

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 \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Ta có: \(VT=\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7bk^2+3bkb}{11bk^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)

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

\(\Rightarrow VT=VP\)

Vậy \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)

Nâng cao r 

mk chịu :)

* VT là vế trái // VP là vế phải *

\(#tutuuu..\)