Question

Chứng minh: Nếu \(\frac{a}{b}=\frac{c}{d}\)

Thì : a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)

        b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)

Answer 1

a)Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\Leftrightarrow\frac{5bk+3b}{5bk-3b}=\frac{5dk+3d}{5dk-3d}\)

Xét VT \(\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(1\right)\)

Xét VP \(\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)

Từ (1) và (2) -->Đpcm

b)Đặt tương tự ta xét VT

\(\frac{7\left(bk\right)^2+3bk\cdot b}{11\left(bk\right)^2-8b^2}=\frac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(1\right)\)

Xét VP \(\frac{7\left(dk\right)^2+3dk\cdot d}{11\left(dk\right)^2-8d^2}=\frac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(2\right)\)

Từ (1) và (2) -->Đpcm

Answer 2

ờ đề hoj j nx