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Ta có:
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+d}=\dfrac{a.b}{c.d}\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a}{b}=\dfrac{c}{d}\)
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:\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
CHÚC BẠN HỌC TỐT!!
\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)( 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 \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Ta đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=k\) => a=ck ; b=dk
a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{c^2-d^2}=\dfrac{b^2k^2-d^2k^2}{c^2-d^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\)(1)
\(\dfrac{ab}{cd}=\dfrac{ck.dk}{cd}=\dfrac{k^2\left(c.d\right)}{cd}=k^2\) (2)
Từ (1) và (2) => \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)
b) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(ck-dk\right)^2}{\left(c-d\right)^2}=\dfrac{k^2\left(c-d\right)^2}{\left(c-d\right)^2}=k^2\) (3)
Từ (2) và (3) => \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\). Chúc bạn học tốt 
Ta có :\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow cd\left(a^2+b^2\right)=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=c^2ab+d^2ab\)
\(\Leftrightarrow\left(a^2cd+b^2cd\right)-\left(c^2ab+d^2ab\right)=0\)
\(\Leftrightarrow aacd+bbcd-ccab-ddab=0\)(tất cả là dấu nhân ko phải số tự nhiên có 4 chữ số nha)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ad-bc\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ad-bc=0\\ac-bd=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\left(\text{đ}pcm\right)\)
bn ơi cho mk hỏi ac vói bd rút gọn kiểu gì mà nó mất đc
Ta đặt : a/b = c/d = K ( K khác 0 )
=> a = b.K
c = d.K
Mà : a2 + b2 / c2 + d2 = b.K2 + b2 / d.K2 + d2
= b2 . ( K2 + 1 ) / d2 . ( K2 + 1 )
= b2 / d2 ( 1 )
Mà : ab/cd = b.K.b / d.K.d = b2 . K / d2 . K
= b2 / d2 ( 2 )
Từ ( 1 ) và ( 2 ) suy ra : a/b =c/d ( ĐPCM )
ta có \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\Rightarrow ab.\left(c^2+d^2\right)=cd.\left(a^2+b^2\right)\)
suy ra \(ab.\left(c^2+d^2\right)\)=\(abc^2+abd^2=acbc+adbd\) (1)
\(cd\left(a^2+b^2\right)=a^2cd+b^2cd+bcbd\) =acad+bcbd (2)
(1);(2) suy ra acbc+adbd=acad+bcbd
nên bc+ad=bc+ad
suy ra ad=bc nên \(\dfrac{a}{b}=\dfrac{c}{d}\)
Có : \(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số ...... :
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\Leftrightarrow\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{a^2+b^2}{c^2+d^2}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\left(\dfrac{a}{c}^2\right)=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{ab}{cd}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\left(đpcm\right)\)
không biết có đúng không.
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{2ab}{2cd}=\dfrac{a^2+b^2+2ab}{c^2+d^2+2cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\left(1\right)\)
Từ điều 1 ta có:\(\dfrac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}=\dfrac{a.b}{c.d}\Rightarrow\dfrac{c.\left(a+b\right)}{a.\left(c+d\right)}=\dfrac{b.\left(c+d\right)}{d.\left(a+b\right)}\)
\(\Rightarrow\dfrac{ca+cb}{ca+ad}=\dfrac{cb+bd}{ad+bd}=\dfrac{ca-bd}{ca+bd}=\dfrac{1}{1}\)
\(\Rightarrow ca+cb=ca+ad\Rightarrow cb=ad\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)(đpcm)
Chúc bạn học tốt!!!