BIẾT RẰNG :a^2 +b^2 / c^2+d^2 =ab / cd.Với a,b,c,d khac 0.CMR :
a/b c/d
: biết a^2+b^2/c^2+d^2=ab/cd với a,b,c, d khac 0 Chứng minh rằng :
a/b=c/d hoặc a/b=d/c
cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) voi a; b; c khac 0 va c khac cong tru d . CMR \(\frac{a}{b}=\frac{c}{d}\)
co ai biet ko? Neu biet thi giup mk voi
Cho: a^2+b^2/c^2+d^2=ab/cd (voi a,b,c,d khác 0;c khac +d,-d)
Chứng ming rằng a/b=c/d hoặc a/b=d/c
cho a b c d khac 0 a+b=c+d a^2+b^2=c^2+d^2 cmr a^2006+b^2006=c^2006+d^2006
\(a^2+b^2=c^2+d^2\Leftrightarrow a^2-c^2=d^2-b^2\Leftrightarrow\left(a-c\right)\left(a+c\right)=\left(d-b\right)\left(d+b\right)\)
mà a+b=c+d <=> a-c=d-b <=> \(\left(a-c\right)\left(a+c\right)=\left(a-c\right)\left(d+b\right)\)
TH1: a-c\(\ne0\)<=>a+c=d+b<=>a-b=d-c cộng vế với vế với a+b=c+d (gt) <=> 2a=2d <=> a=d <=> b=c
=>a2006=d2006;b2006=c2006=>a2006+b2006=c2006+d2006
TH2: a-c=0 <=> a=c <=> b=d <=> a2006+b2006=c2006+d2006
Từ 2 trường hợp trên suy ra đpcm
cho a/b=c/d khac 1 va a,b,c,d khac 0. chung minh (a-b)^2/(c-d)^2=ab/cd
Cho 4 so a,b,c,d khac 0 thoa man;b^2=ac,c^2=bd,b^3+c^3+d^3 khac 0
CMR;a^3+b^3+c^3/b^3+c^3+d^3=a/d
Cho a/b=c/d và b+d khac 0.CMR: 3a2+c2/3b2+d2=(a+c)2/(b+d)2
Đặt\(\frac{a}{b}=\frac{c}{d}=k\left(k\in Q\right)\)
\(\Rightarrow\hept{\begin{cases}a=bk\left(1\right)\\c=dk\left(2\right)\end{cases}}\)
Ta lại có \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(3\right)\)
Thay \(\left(1\right),\left(2\right)vào\left(3\right)có\)
\(\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(4\right)\)
\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(5\right)\)
Từ \(\left(4\right),\left(5\right)\Rightarrowđpcm\)
Cho a^2+b^2tat ca/c^2+d^2 =ab/cd
va a,b,c,d khac 0
cm a/b=c/d hoac a/b=d/c
cho a/b=c/d khac 1 va c khac 0
CMR:
a)((a.b)/(c.d))^2=(a.b)/(c-d)
b)((a.b/c.d))^3=((a^3-b^3)/(a^3-d^3))