Cho tỉ lệ thức a/b=c/d. CM các tỉ lệ thức sau:
a;a^2-b2/ab=c^2-d^2/cd
b;(a+b^2)/a^2+b^2=a^3+b^3/c^3+d^3
K
Khách
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VA
1
MN
8 tháng 1 2016
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\left(\frac{a+b}{c+d}\right)^3=\left(\frac{bk+b}{dk+d}\right)^3=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^3=\left(\frac{b}{d}\right)^3\left(1\right)\)
\(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}=\left(\frac{b}{d}\right)^3\left(2\right)\)
Từ (1) & (2)=>\(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\)
HQ
Hà Quang Minh
Giáo viên
20 tháng 9 2023
a) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)
Ta có \(\dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\)\( \Rightarrow d(a + b) = b(c + d)\)\( \Rightarrow ad + bd = bc + bd\)
\( \Rightarrow ad = bc\) (luôn đúng)
\( \Rightarrow \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\)
b) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)
Ta có: \(\dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}\)
\(\begin{array}{l} \Rightarrow d(a - b) = b(c - d)\\ \Leftrightarrow ad - bd = bc - bd\\ \Leftrightarrow ad = bc\end{array}\) ( luôn đúng)
Vậy \(\dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}\)
c) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)
Ta có: \(\dfrac{a}{{a + b}} = \dfrac{c}{{c + d}}\)
\(\begin{array}{l} \Rightarrow a(c + d) = c(a + b)\\ \Leftrightarrow ac + ad = ac + bc\\ \Leftrightarrow ad = bc\end{array}\) (luôn đúng)
Vậy \(\dfrac{a}{{a + b}} = \dfrac{c}{{c + d}}\)
NN
1
14 tháng 9 2017
vi ab = cd
=>a/b=c/d
=>a+c/b+d =a/b = c/d
=>a-c/b-d =a/b = c/d
(sgk s8 )
22 tháng 8 2021
a) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) và \(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
22 tháng 8 2021
b) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\\\dfrac{c-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\end{matrix}\right.\)\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
d) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{c}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
D
1
19 tháng 1
\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
=>\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
T
2
3 tháng 10 2017
\(\frac{a}{c}\)= \(\frac{d}{b}\); \(\frac{c}{a}\)=\(\frac{b}{d}\);\(\frac{b}{c}\)=\(\frac{d}{a}\); \(\frac{a}{d}\)=\(\frac{c}{b}\)
NK
1
15 tháng 12 2016
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)
\(\Rightarrow bd-ad=bd-bc\)
\(\Rightarrow d\left(b-a\right)=b\left(d-c\right)\)
\(\Rightarrow\frac{b-a}{b}=\frac{d-c}{d}\left(đpcm\right)\)
14 tháng 10 2021
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
14 tháng 10 2021
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a^2-b^2}{ab}=\frac{\left(bk\right)^2-b^2}{bk.b}=\frac{b^2.k^2-b^2}{b^2k}=\frac{b^2\left(k^2-1\right)}{b^2k}=\frac{k^2-1}{k}\left(1\right)\)
\(\frac{c^2-d^2}{cd}=\frac{\left(dk\right)^2-d^2}{dk.d}=\frac{d^2k^2-d^2}{d^2k}=\frac{d^2\left(k^2-1\right)}{d^2.k}=\frac{k^2-1}{k}\left(2\right)\)
Từ (1) và (2)=>\(\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\).
phần b đề kiểu gì vậy??//