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1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\hept{\begin{cases}\frac{a^2}{c^2}=\frac{b^2}{d^2}\\\frac{a^2}{c^2}=\frac{ab}{cd}\end{cases}}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\) (*)
ADTCDTSBN
có: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
Từ (*) \(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\) ( đ p c m)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)\cdot cd=\left(c^2+d^2\right)\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd=c^2\cdot ab+d^2\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd-c^2\cdot ab-d^2\cdot ab=0\)
\(\Rightarrow\left(a^2\cdot cd-c^2\cdot ab\right)+\left(b^2\cdot cd-d^2\cdot ab\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)+bd\cdot\left(bc-ad\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)-bd\cdot\left(ad-bc\right)=0\)
\(\Rightarrow\left(ac-bd\right)\cdot\left(ad-bc\right)=0\)
Tự làm tiếp nhé.......
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right).cd=ab.\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abd^2=abc^2-b^2cd\)
\(\Leftrightarrow ad\left(ac-bd\right)=bc\left(ac-bd\right)\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
Ta có:
\(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Ta có:
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{ab}{dc}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1)(2) => \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
a) => \(\frac{2a+c}{2b+d}=\frac{2kb+kd}{2b+d}=\frac{k\left(2b+d\right)}{2b+d}=k\) (1)
\(\frac{2a-3c}{2b-3d}=\frac{2kb-3kd}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\) (2)
Từ (1) và (2) => \(\frac{2a+c}{2b+d}=\frac{2a-3c}{2b-3d}\)
b) => \(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) => \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
a) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
Đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)= k ( k \(\in\)Z , k khác 0 )
=> a = bk ; c = dk
Ta có:
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
Vậy nếu \(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
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