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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: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)\cdot cd=\left(c^2+d^2\right)\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd=c^2\cdot ab+d^2\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd-c^2\cdot ab-d^2\cdot ab=0\)
\(\Rightarrow\left(a^2\cdot cd-c^2\cdot ab\right)+\left(b^2\cdot cd-d^2\cdot ab\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)+bd\cdot\left(bc-ad\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)-bd\cdot\left(ad-bc\right)=0\)
\(\Rightarrow\left(ac-bd\right)\cdot\left(ad-bc\right)=0\)
Tự làm tiếp nhé.......
Giải:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\)
a, Ta có: \(k^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\) (1)
\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, Ta có: \(k=\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow k^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (1)
\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)Thay vào \(\frac{a^2-b^2}{c^2-d^2}\) ta được:
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{b^2k^2-b^2}{d^2k^2-d^2}\Rightarrow\frac{b^2}{d^2}\Rightarrow\frac{b.b}{d.d}\left(1\right)\)
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow a=b;c=d\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k=> a=bk; c=dk
vế trái =\(\frac{ab}{cd}\)=\(\frac{bk.b}{dk.d}\)=\(\frac{b^2}{d^2}\)(1)
vế phải =\(\frac{\left(a+b\right)2}{\left(c+d\right)^2}\)=\(\frac{\left(bk+b\right)2}{\left(dk+d\right)^2}\)=\(\frac{b^2\left(k+1\right)2}{d^2\left(k+1\right)^2}\)=\(\frac{b^2}{d^2}\)(2)
Từ (1) và (2) ta có:\(\frac{ab}{cd}\)=\(\frac{\left(a+b\right)2}{\left(c+d\right)^2}\)
a
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)
b
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
c
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{5a^2}{5b^2}=\frac{3c^2}{3d^2}=\frac{5a^2+3c^2}{3d^2+5b^2}\)
b) Ta có: \(a.d=b.c\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
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) => \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
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)