cho a/b=c/d chứng minh rằng:
a)\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\) b)\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
cho \(\dfrac{a^2+b^2}{c^2+d^2}\)= \(\dfrac{ab}{cd}\).Chứng minh rằng: hoặc \(\dfrac{a}{b}\)= \(\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}\)= \(\dfrac{d}{c}\)
cho\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)chứng minh rằng \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\end{matrix}\right.\\ \RightarrowĐpcm\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh rằng:
a) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b)\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c)\(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
a; Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7b^2k^2-3\cdot bk\cdot b}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2-3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2-3k}{11k^2-8}\)
\(\dfrac{7c^2-3cd}{11c^2-8d^2}=\dfrac{7d^2k^2-3kd^2}{11d^2k^2-8d^2}=\dfrac{7k^2-3k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2-3cd}{11c^2-8d^2}\)
Cho các số a,b,c,d thỏa mãn \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}.\) Chứng minh rằng: ad=bc hoặc ac=bd
theo bài ra ta có:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\\ \Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{2ab}{2cd}\)
áp dụng tính chất dãy tỉ số bàng nhau ta có:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{2ab}{2cd}=\dfrac{a^2+b^2+2ab}{c^2+d^2+2cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\\ \Rightarrow\dfrac{c\left(a+b\right)}{a\left(c+d\right)}=\dfrac{b\left(c+d\right)}{d\left(a+b\right)}\\ \Rightarrow\dfrac{ca+cb}{ca+ad}=\dfrac{bc+bd}{ad+bd}\)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{ca+cb}{ca+ad}=\dfrac{bc+bd}{ad+bd}=\dfrac{\left(ca+cb\right)-\left(bc+bd\right)}{\left(ca+ad\right)-\left(ad+bd\right)}=\dfrac{ca-bd}{ca-bd}=1\\ \Rightarrow ca+cb=ca+ad\\ \Rightarrow cb=ad\\ \Rightarrow ad=bc\left(đpcm\right)\)
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng
a) \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
a, Ta có: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)
\(\Rightarrow\dfrac{b^2.k}{d^2.k}=\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b, Ta có:\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
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\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)( áp dụng tỉ lệ thức )
Ta đặt:
\(\dfrac{a}{c}=\dfrac{b}{d}=k\) => a=ck ; b=dk
a) \(\dfrac{ab}{cd}=\dfrac{ck.dk}{cd}=\dfrac{k^2.\left(c.d\right)}{c.d}=k^2\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(ck+dk\right)^2}{\left(c+d\right)^2}=\dfrac{k^2.\left(c+d\right)^2}{\left(c+d\right)^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(ck\right)^2+\left(dk\right)^2}{c^2+d^2}=\dfrac{c^2k^2+d^2k^2}{c^2+d^2}=\dfrac{k^2.\left(c^2+d^2\right)}{c^2+d^2}=k^2\) (3)
Từ (1) và (3) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
ab=cdab=cd=>ac=bdac=bd( áp dụng tỉ lệ thức )
Ta đặt:
ac=bd=kac=bd=k => a=ck ; b=dk
a) abcd=ck.dkcd=k2.(c.d)c.d=k2abcd=ck.dkcd=k2.(c.d)c.d=k2 (1)
(a+b)2(c+d)2=(ck+dk)2(c+d)2=k2.(c+d)2(c+d)2=k2(a+b)2(c+d)2=(ck+dk)2(c+d)2=k2.(c+d)2(c+d)2=k2 (2)
Từ (1) và (2) suy ra abcd=(a+b)2(c+d)2abcd=(a+b)2(c+d)2
b) a2+b2c2+d2=(ck)2+(dk)2c2+d2=c2k2+d2k2c2+d2=k2.(c2+d2)c2+d2=k2a2+b2c2+d2=(ck)2+(dk)2c2+d2=c2k2+d2k2c2+d2=k2.(c2+d2)c2+d2=k2 (3)
Từ (1) và (3) suy ra abcd=a2+b2c2+d2
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)(b, c, d ≠ 0 , b + d ≠ 0). Chứng minh rằng: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Theo đề bài ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) ( 1 )
Theo tính chất dãy tỉ số bằng nhau ta có :
\(k=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(k^2=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( 2 )
Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )
Từ ( 2 ) , ( 3 )
= > \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( đpcm )
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng ta có các tỉ lệ thức sau (giả thiết các tỉ lệ thức là có nghĩa ) :
a) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
b) \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
cho\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\).Chứng minh rằng \(\dfrac{ab}{cd}\)= \(\dfrac{a^2-b^2}{c^2-d^2}\).Mình đang cần gấp ạ, mong mọi người giúp mình!