Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
CMR \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
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NT
3 tháng 1 2018
\(\dfrac{a}{b}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\\ \Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\\ \dfrac{a^2}{c^2}=\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
4 tháng 1 2018
Có thể dùng cách khác:v
a)\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=t\)(với t là 1 số thực bất kì thỏa mãn)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)
Tương tự:v
24 tháng 10 2021
d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)
\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)
Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
24 tháng 5 2022
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk\cdot b}=\dfrac{b^2\left(k^2-1\right)}{b^2k}=\dfrac{k^2-1}{k}\)
\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk\cdot d}=\dfrac{d^2\left(k^2-1\right)}{d^2\cdot k}=\dfrac{k^2-1}{k}\)
Do đó: \(\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{b^2\cdot\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\)
\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left(k+1\right)^2}{k^2+1}\)
Do đó: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
25 tháng 5 2022
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk và c=dk
ta có \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)\(\dfrac{ab}{cd}=\dfrac{bk.b}{bk.d}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) (cùng =\(\dfrac{b^2}{d^2}\) ) (đpcm)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=t\)
a) \(\left\{{}\begin{matrix}\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\end{matrix}\right.\Rightarrowđpcm\)
b) \(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\left(\dfrac{a+b}{c+d}\right)^2=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)