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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.
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:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a+b+c}{b+c+d}\\\dfrac{b}{c}=\dfrac{a+b+c}{b+c+d}\\\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\) (đpcm)
bn cũng có thể tham khảo
https://hoc24.vn/hoi-dap/question/466226.html
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Thay vào ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2\cdot k}{d^2\cdot k}=\dfrac{b^2}{d^2}\left(1\right)\)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)
\(=\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\)
\(=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) suy ra: đpcm
Gia su \(\dfrac{a}{b}=\dfrac{c}{d}=k\)=> a=bk; c=dk
The vao ta co:
\(\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)<=>\(\dfrac{b^2\cdot k}{d^2\cdot k}=\dfrac{b^2\cdot k^2-b^2}{d^2\cdot k^2-d^2}\)<=>\(\dfrac{b^2}{d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}\)
=>\(\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ⇒ a=bk, c=dk
a) Ta có: ✽ \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
✽\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
nên \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{a-c}{c}=\dfrac{bk-dk}{dk}=\dfrac{k\left(b-d\right)}{dk}=\dfrac{b-d}{d}\)
Vậy \(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\) =>\(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
=>\(\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\) =>\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)(đpcm)
Chúc Bạn học Tốt
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Ta đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=k\) => a=ck ; b=dk
a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{c^2-d^2}=\dfrac{b^2k^2-d^2k^2}{c^2-d^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\)(1)
\(\dfrac{ab}{cd}=\dfrac{ck.dk}{cd}=\dfrac{k^2\left(c.d\right)}{cd}=k^2\) (2)
Từ (1) và (2) => \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)
b) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(ck-dk\right)^2}{\left(c-d\right)^2}=\dfrac{k^2\left(c-d\right)^2}{\left(c-d\right)^2}=k^2\) (3)
Từ (2) và (3) => \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\). Chúc bạn học tốt 
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!!
Câu hỏi tương tự
1, Từ a/b=b/c=c/d => (a/d)3= (b/c)3=(c/d)3= a.b.c/b.c.d=a/d (1)
Lại có: a/b = b/c= c/d = a+b+c/b+c+d= (a/d)3= (b/c)3=(c/d)3=(a+b+c/b+c+d)3 (2)
Tưd 1 và 2 suy ra ..... (đpcm)
2, Từ a/b = b/c= c/d = a+b+c/b+c+d=1
=> a=b=c