Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\left(đpcm\right)\)
b;c;d tương tự hết
b: a/b=c/d
nên 3a/3b=2c/2d
=>a/b=c/d=(3a+2c)/(3b+2d)
c: a/c=b/d nên a/c=2b/2d=(a-2b)/(c-2d)
d: a/c=b/d
nên 5a/5c=2b/2d
=>a/c=b/d=(5a-2b)/(5c-2d)
a/ Ta có \(a\left(2a-5c\right)=2a^2-5ac=2bc-5ac=c\left(2b-5a\right)\Rightarrow\frac{c}{2a-5c}=\frac{a}{2b-5a}\)
Các câu khác làm tương tự
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
ta có:\(\dfrac{5a+3b}{7a-2b}=\dfrac{5.ck+3.dk}{7.ck-2.dk}=\dfrac{k.\left(5c+3d\right)}{k.\left(7c-2d\right)}=\dfrac{5c+3d}{7c-2d}\)Vậy \(\dfrac{5a+3b}{7a-2b}=\dfrac{5c+3d}{7c-2d}\left(đpcm\right)\)
b) từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
ta có:\(\dfrac{7a^2+3ab}{11a^2+8.b^2}=\dfrac{7.c^2.k^2+3.c.d.k^2}{11.c^2.k^2+8.d^2.k^2}=\dfrac{k^2.\left(7.c^2+3.c.d\right)}{k^{2.}\left(11.c^2+8.d^2\right)}\) vậy .......
c)\(từ\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\dfrac{a+b}{c+d}\right)^2\)(1)
Mặt khác:\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1).(2)=>......
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
<=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{3a}{3c}=\dfrac{2b}{2d}\)
<=>\(\dfrac{5a-3b}{5c-3d}=\dfrac{3a-2b}{3c-2d}\)(đpcm)
Các câu sau tương tự
a/a+2b=c/c+2d => a.(c+2d)=c.(a+2b)
=> ac+2da=ac+2bc
=> 2da=2bc
=>da=bc
=> a^2.d^2-4b^2.c^2/abcd=ad.ad-4.bc.bc/ad.bc=bc.bc-4.bc.bc/bc.bc=bc.bc.(1-4)=bc.bc=1-4/1=-3/1=-3