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Câu 1
Ta có : \(\frac{a}{b}=\frac{c}{d}=>\left(\frac{a}{b}+1\right)=\left(\frac{c}{d}+1\right)\left(=\right)\frac{a+b}{b}=\frac{c+d}{d}\)
=> ĐPCM
Câu 2
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{b}{a}=\frac{d}{c}=>\left(\frac{b}{a}+1\right)=\left(\frac{d}{c}+1\right)\left(=\right)\frac{b+a}{a}=\frac{d+c}{c}=>\frac{a}{b+a}=\frac{c}{d+c}\)
=> ĐPCM
Câu 3
Câu 3
Ta có \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(=) (a+b).(c-d)=(a-b).(c+d)(=)ac-ad+bc-bd=ac+ad-bc-bd(=)-ad+bc=ad-bc(=) bc+bc=ad+ad(=)2bc=2ad(=)bc=ad=> \(\frac{a}{b}=\frac{c}{d}\)
=> ĐPCM
Câu 4
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=>\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\left(1\right)\)
Lại có \(\frac{a^2+c^2}{b^2+d^2}=\frac{b^2k^2+c^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2) => ĐPCM
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk ; c=dk
=>\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\left(1\right)\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{\left(b^2\cdot k^2\right)+b^2}{\left(d^2\cdot k^2\right)+d^2}=\dfrac{b^2\cdot\left(k^2+1\right)}{d^2\cdot\left(k^2+1\right)}=\dfrac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) =>\(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(VT=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(ck\right)^2+\left(dk\right)^2}{c^2+d^2}=\dfrac{c^2.k^2+d^2.k^2}{c^2+d^2}=\dfrac{k^2.\left(c^2+d^2\right)}{c^2+d^2}=k^2\left(1\right)\)
\(VP=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(ck+dk\right)^2}{\left(c+d\right)^2}=\dfrac{\left(ck\right)^2+2.ck.dk+\left(dk\right)^2}{c^2+2.c.d+d^2}=\dfrac{c^2.k^2+2.c.d.k^2+d^2.k^2}{c^2+2.c.d+d^2}=\dfrac{k^2.\left(c^2+2.c.d+d^2\right)}{c^2+2.c.d+d^2}=k^2\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(đpcm\right)\)
Vậy ...
Đặt \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k
\(\dfrac{a}{b}\)= k => a = bk
\(\dfrac{c}{d}\)= k => c = dk
khi đó
\(\dfrac{a^2+b^2}{c^2+d^2}\)= \(\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)=\(\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\)=\(\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)= \(\dfrac{b^2}{d^2}\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)=\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)=\(\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)=\(\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\)=\(\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)= \(\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) => \(\dfrac{a^2+b^2}{c^2+d^2}\)= \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
mk mk làm cho