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Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk\),\(c=dk\)
\(\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\left(1\right)\)
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)(đpcm)
Đặt \(\dfrac{a}{b}=k;\dfrac{c}{d}=k\)
\(\Rightarrow a=kb;c=kd\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=k^2\)
Từ các chứng minh trên cho ta thấy
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a.c}{b.d}\)
cho \(b^2=a.c-a^2=b.d\)
c/m:\(\dfrac{a^3+b^3-c^3}{b^3+c^2-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^2\)
\(b^2=a.c\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=b.d\)
\(\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
\(=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}\left(đpcm\right)\)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)
Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
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)
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(c^2=\dfrac{b}{c}=\dfrac{c}{d}\)
Do đó: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Do đó: \(\dfrac{a^3.b^3.c^3}{b^3.c^3.d^3}=\dfrac{a}{d}\left(đpcm\right)\)
Vậy ...............
Chúc bạn học tốt!
a: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)
b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k=\dfrac{a}{b}\)
c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)
d: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\\ \dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)