Question

cho \(\frac{a}{b}=\frac{c}{d}\): CMR :

\(a,\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\)

\(b,\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)

Answer 1

nhân chéo r rút gọn

Answer 2

Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{4a}{4c}=\frac{3b}{3d}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\)

Vậy \(\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\left(ĐPCM\right)\)

Answer 3

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\left(k\ne0\right)\)

a) \(\frac{4a-3b}{4c-3d}=\frac{4bk-3b}{4dk-3d}=\frac{b.\left(4k-3\right)}{d.\left(4k-3\right)}=\frac{b}{d}\)

      \(\frac{4a+3b}{4c+3d}=\frac{4bk+3b}{4dk+3d}=\frac{b.\left(4k+3\right)}{d.\left(4k+3\right)}=\frac{b}{d}\)

\(\Rightarrow\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\)

b) \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3.\left(k^3+1\right)}{d^3.\left(k^3+1\right)}=\frac{b^3}{d^3}\)

     \(\frac{a^3-b^3}{c^3-d^3}=\frac{\left(bk\right)^3-b^3}{\left(dk\right)^3-d^3}=\frac{b^3k^3-b^3}{d^3k^3-d^3}=\frac{b^3.\left(k^3-1\right)}{d^3.\left(k^3-1\right)}=\frac{b^3}{d^3}\)

\(\Rightarrow\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)