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DL
1
NL
3
20 tháng 6 2017
Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
20 tháng 6 2017
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\left(1\right)\)
a) Thay (1) vào đề:
\(VT=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)
\(VP=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\)
\(\Rightarrow VT=VP\)
\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)
b) Thay (1) vào đề bài:
\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)
Theo câu a) \(\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)
2 tháng 11 2016
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
a) => \(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{kb-b}{kd-d}\right)^2=\left(\frac{b\left(k-1\right)}{d\left(k-1\right)}\right)^2=\left(\frac{b}{d}\right)^2\) (1)
\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) => \(\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\)
b)=> \(\left(\frac{a+b}{c+d}\right)^3=\left(\frac{kb+b}{kd+d}\right)^3=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^3=\frac{b^3}{d^3}\) (1)
\(\frac{a^3-b^3}{c^3-d^3}=\frac{\left(kb\right)^3-b^3}{\left(kd\right)^2-d^3}=\frac{b^3\left(k^3-1\right)}{d^3\left(k^3-1\right)}=\frac{b^3}{d^3}\) (2)
Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\)
P
0
Đặt a/b=c/d=k
=>a=kb
c=kd
Ta có:\(\frac{a}{b-a}=\frac{kb}{b-kb}=\frac{kb}{b\left(k-1\right)}=\frac{k}{k-1}\)
\(\frac{c}{d-c}=\frac{kd}{d-kd}=\frac{kd}{d\left(k-1\right)}=\frac{k}{k-1}\)
=>\(\frac{a}{b-a}=\frac{c}{d-c}\)
Ta có:\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{k^2b^2+b^2}{k^2d^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)
\(\frac{ab}{cd}=\frac{kb.b}{kd.d}=\frac{b^2}{d^2}\)
=>\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\frac{a}{b}=\frac{c}{d}=>\frac{b}{a}=\frac{d}{c}=>\frac{b}{a}-1=\frac{d}{c}-1=>\frac{b-a}{a}=\frac{d-c}{c}=>\frac{a}{b-a}=\frac{c}{d-c}\)
=>ĐPCM
\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{a}{c}.\frac{b}{d}=\frac{b}{d}.\frac{b}{d}=>\frac{ab}{cd}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{a}{c}=>\frac{a^2}{c^2}=\frac{ab}{cd}\left(2\right)\)
Từ (1) và (2) ta thấy:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
=>\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
=>ĐPCM