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Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{b}{d}=\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)
\(\Leftrightarrow\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)
đpcm
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(=>\hept{\begin{cases}a=b.k\\c=d.k\end{cases}}\)
\(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{b.k-b}{d.k-d}\right)^2=\left(\frac{b.\left(k-1\right)}{d.\left(k-1\right)}\right)^2\)\(=\frac{\left(b^2.\left(k-1\right)^2\right)}{\left(d^2.\left(k-1\right)^2\right)}=\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}=\frac{b^2}{d^2}\)\(\left(1\right)\)
\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\)
Đặt \(\frac{a}{b}\)= \(\frac{c}{d}\)= k => a= bk ; c = dk
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)= \(\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}\)= \(\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}\)= \(\frac{bk.b}{dk.d}\)= \(\frac{b^2}{d^2}\) (2)
Từ (1) và (2) ->> \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{ab}{cd}\)
a) sai đề rồi bn
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\left(\frac{a+b}{c+d}\right)^3\)(tính chất dãy tỉ số bằng nhau) (1)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3-b^3}{c^3-d^3}\)(2)
từ (1) và (2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\left(đpcm\right)\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
đặt: \(\frac{a}{c}=\frac{b}{d}=t\) Áp dụng Tính chất dãy tỉ số bằng nhau:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=t\Leftrightarrow\left(\frac{a-b}{c-d}\right)^2=t^2\)
\(\frac{a}{c}=\frac{b}{d}=t\Leftrightarrow\frac{ab}{cd}=t^2\)
\(\Rightarrowđpcm\)
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.......
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ca+bc\)
\(\Rightarrow ab-cb=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Trả lời :........................................................
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}......................\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Học sinh giỏi 6A
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: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)