Cho tỉ lệ thức a/b =c/d Chứng minh rằng:
a)a-b/a = c-d/c
b)a+b/a-b=c+d/c-d
c)(a-b/c-d)2 = ab/cd
d)(a+b/c+d)3 = 33-b3/c3-d3
e)2a+3b/2a-3b = 2c+3d/2c-3d
g)ab/cd = a2-b2/c2-d2
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c)\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(ck+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
k cho mình nhé
Sửa đề: Chứng minh \(\dfrac{ab}{cd}=\left(\dfrac{2a+3b}{2c+3d}\right)^2\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{2a+3b}{2c+3d}\right)^2=\left(\dfrac{2bk+3b}{2dk+3d}\right)^2=\left(\dfrac{b}{d}\right)^2\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{2a+3b}{2c+3d}\right)^2\)
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b}{d}\)
\(\dfrac{2a-3b}{2c-3d}=\dfrac{2bk-3b}{2dk-3d}=\dfrac{b}{d}\)
Do đó: \(\dfrac{a-b}{c-d}=\dfrac{2a-3b}{2c-3d}\)