chứng minh tỉ lệ thức
\(\dfrac{a.b}{c.d}\)=\(\dfrac{a^2-b^2}{c^2-d^2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2021
Đẳng thức đầu tiên sai:
Ví dụ: \(a=1;b=2;c=3;d=6\) thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Nhưng \(\dfrac{a.d}{c.d}\ne\dfrac{a^2-b^2}{b^2-d^2}\)
Với đẳng thức thứ 2:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
LN
0
28 tháng 10 2016
Ta có:
\(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}=\frac{a^2+b^2+a.b}{c^2+d^2+c.d}=\frac{a^2+a.b+b^2+a.b}{c^2+c.d+d^2+c.d}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}=\frac{a\left(a+b\right)+b\left(a+b\right)}{c\left(c+d\right)+d\left(c+d\right)}=\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\)
\(\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}=\frac{a.b}{c.d}\Rightarrow\frac{c\left(a+b\right)}{a\left(c+d\right)}=\frac{b\left(c+d\right)}{d\left(a+b\right)}\)
\(\Rightarrow\frac{ca+cb}{ca+ad}=\frac{bc+bd}{ad+bd}=\frac{ca+bd}{ca-bd}=1\)
\(\Rightarrow ca+cb=ca+ad\)
\(\Rightarrow cb=ad\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
M
2
D
11 tháng 10 2017
Ta có tỉ lệ thức
\(\frac{a}{b}=\frac{c}{d}\)
Suy ra
a=bk
c=dk
Nên ta có
\(\frac{a.b}{c.d}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2} \)
Suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
12 tháng 11 2016
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{a^2}{b^2};\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}=\frac{c^2}{d^2}\\ \Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
DM
26 tháng 6 2017
Ta có:
\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{a.b}{c.d}\)=\(\frac{a^2+b^2+a.b}{c^2+d^2+c.d}\)=\(\frac{a^2+a.b+b^2+a.b}{c^2+c.d+d^2+c.d}\)
\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{a.b}{c.d}\)=\(\frac{a\left(a+b\right)+b\left(a+b\right)}{c\left(c+d\right)+d\left(c+d\right)}\)\(\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\)
\(\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\)=\(\frac{a.b}{c.d}\)=) \(\frac{c\left(a+b\right)}{a\left(c+d\right)}\)=\(\frac{b\left(c+d\right)}{d\left(a+b\right)}\)
=) \(\frac{ca+cb}{ca+ad}\)=\(\frac{bc+bd}{ad+bd}\)=\(\frac{ca-bd}{ad-bd}\)=1
=) ca + cb = ca + ad
=) cb = ad
=) \(\frac{a}{b}\)= \(\frac{c}{d}\)
13 tháng 12 2016
a) Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2.\left(k^2-1\right)}{d^2.\left(k^2-1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
b) Giải:
Để \(P\in Z\Rightarrow2x-3⋮x+1\)
Ta có:
\(2x-3⋮x+1\)
\(\Rightarrow\left(2x+2\right)-5⋮x+1\)
\(\Rightarrow5⋮x+1\)
\(\Rightarrow x+1\in\left\{1;-1;5;-5\right\}\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
+) \(x+1=5\Rightarrow x=4\)
+) \(x+1=-5\Rightarrow x=-6\)
Vậy \(x\in\left\{0;-2;4;-6\right\}\)
\(\Rightarrow5⋮x+1\)
13 tháng 12 2016
1)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)(tính chất dãy tỉ số bằng nhau)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
2)\(P=\frac{2x-3}{x+1}=\frac{2x+2-5}{x+1}=\frac{2\left(x+1\right)-5}{x+1}=2-\frac{5}{x+1}\)
\(\Rightarrow P\in Z\Leftrightarrow2-\frac{5}{x+1}\in Z\Leftrightarrow\frac{5}{x+1}\in Z\Leftrightarrow5⋮x+1\Leftrightarrow x+1\inƯ\left(5\right)\)
\(\Rightarrow x+1\in\left\{-1;-5;1;5\right\}\)
\(\Rightarrow x\in\left\{-2;-6;0;4\right\}\)
2 tháng 9 2017
Theo đề bài, ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\left(\dfrac{a+b}{c+d}\right)^2\)(*)
=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{a^2+b^2}{c^2+d^2}\)(**)
Từ (*) và (**) suy ra:
\(\left(\dfrac{a+b}{c+d}\right)^2\)=\(\dfrac{a^2+b^2}{c^2+d^2}\)(đpcm)
Đặt a/b=c/d=k
suy ra a=bk
b=dk
Từ đó ta có: a.b/c.d=bk.b/dk.d=b^2/d^2
a^2-b^2/c^2-d^2= (bk)^2-b^2/(dk)^2-d^2=b^2(k^-1)/d^2(k^2-1)=b^2/d^2
vậy a.b/c.d=a^2-b^2/c^2-d^2(=b^2.d^2)