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Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow \left\{\begin{matrix} a=bk\\ c=dk\end{matrix}\right.\)
Khi đó:
\(\frac{3a^2+c^2}{3b^2+d^2}=\frac{3(bk)^2+(dk)^2}{3b^2+d^2}=\frac{k^2(3b^2+d^2)}{3b^2+d^2}=k^2(1)\)
Và: \(\frac{(a+c)^2}{(b+d)^2}=\frac{(bk+dk)^2}{(b+d)^2}=\frac{k^2(b+d)^2}{(b+d)^2}=k^2(2)\)
Từ \((1); (2)\Rightarrow \frac{3a^2+c^2}{3b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)
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Cảm ơn các bạn nhiều nha
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\) a = bk ; c = dk
\(\Rightarrow\)\(\dfrac{4a^2+4c^2}{4b^2+4d^2}\)=\(\dfrac{4\left(bk\right)^2+4\left(dk\right)^2}{4b^2+4d^2}\)
=\(\dfrac{4b^2k^2+4d^2k^2}{4b^2+4d^2}\)=\(\dfrac{k^2\left(4b^2+4d^2\right)}{4b^2+4d^2}\)= k2 (1)
\(\Rightarrow\)\(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)=\(\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}\)=\(\dfrac{[k\left(b-d\right)]^2}{\left(b-d\right)^2}\)
=\(\dfrac{k^2\left(b-d\right)^2}{\left(b-d\right)^2}\)= k2 (2)
Từ (1) và (2), suy ra:
\(\dfrac{4a^2+4c^2}{4b^2+4d^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\) (đpcm)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*)suy ra:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)\(=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
b) Tương tự câu a nhé bạn!
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
a) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\Rightarrow ad=cb\)
=> \(ad+bd=bc+bd\)
\(\Rightarrow d\left(a+b\right)=b\left(c+d\right)\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
\(\Rightarrow\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\rightarrowđpcm\)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\)
\(\Rightarrow\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ; ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
bạn sửa hộ mik \(\left(\dfrac{a^2+b^2}{c^2+d^2}\right)^2\) thành\(\dfrac{a^2+b^2}{c^2+d^2}\)nha!!
Ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{aa}{bb}=\dfrac{a^2+a^2}{b^2+b^2}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{a^2.2}{b^2.2}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{a^2}{b^2}\)
\(\Leftrightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\rightarrowđpcm\)
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\) (theo đề bài)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
Vậy \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}.\)
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^2=\left(\dfrac{c}{d}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)