đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk\) và \(c=dk\)
thay vào biểu thức
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\) (1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\) (2)
Từ 1 và 2 suy ra đpcm
câu b tương tự bạn thay a=bk và c=dk rồi rút gọn như câu a là xong nha!
Chứng minh nếu \(\frac{a}{b}=\frac{c}{d}\) thì
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
chứng minh rằng :
\(\frac{a}{b}=\frac{c}{d}\)thì
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}\)= \(\frac{7c^2+3cd}{11c^2-8d^2}\)
Chứng minh nếu \(\frac{a}{b}=\frac{c}{d}\) thì :
a, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a, \(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
c, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Bài 1: biết\(\frac{a}{b}=\frac{c}{d}CMR\)
\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR Nếu \(\frac{a}{b}=\frac{c}{d}\)thì:
a)\(\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
b)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
c)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR neu \(\frac{a}{b}=\frac{c}{d}\) thi
a)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR:
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^3+3ab}{11a^3-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho a/b=c/d. Chứng minh:
a: 5a+3b/5a-3b = 5c+3d/5c-3d
b: 7a^2 +3ab/11a^2-8b^2 = 7c^2+3cd/11c^2-8d^2
