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17 tháng 5 2019

Đặt  a b = c d = k ( k ∈ R ) ⇒ a = k . b ; c = k . d

Ta có:  5 a + 3 b 3 a − 7 b = 5 k b + 3 b 3 k b − 7 b = b 5 k + 3 b 3 k − 7 = 5 k + 3 3 k − 7 ( 1 ) 5 c + 3 d 3 c − 7 d = 5 k d + 3 d 3 k d − 7 d = d 5 k + 3 d 3 k − 7 = 5 k + 3 3 k − 7 ( 2 )

Từ (1), (2) => đpcm

14 tháng 7 2016

Chứng minh đẳng thức ad = cd (c,d khác 0) , ta suy ra được tỉ lệ thức a/c =b / d

Hãy lập các tỉ lệ thức từ 4 trong 5 số sau đây

4 ; 16 ; 64 ; 256 ; 1024

1: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=b\cdot k;c=d\cdot k\)

\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

Do đó: \(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

3: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\cdot\left(k-1\right)}=\dfrac{k+1}{k-1}\)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)

Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)

\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

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 \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

28 tháng 10 2018

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:\(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\left(\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}\right)^3=\dfrac{b^3}{d^3}\)(1)

Lại có :\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3.\left(k^3+1\right)}{d^3.\left(k^3+1\right)}=\dfrac{b^3}{d^3}\)(2)

Từ (1) và (2) => ĐPCM

28 tháng 10 2018

Từ a/b=c/d

=>a/c=b/d=a+b/c+d

<=>a^3/c^3=b^3/d^3=(a+b)^3(c+d)^3

=a^3+b^3/c^3+d^3

Vậy

(a+b)^3(c+d)^3=a^3+b^3/c^3+d^3 (đpcm)

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3}{d^3}\)

\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)

Do đó: \(\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\)

25 tháng 7 2017

\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Leftrightarrow\dfrac{a+b+c+d}{a-b+c-d}=\dfrac{a+b-c-d}{a-b-c+d}\)

Theo tính chất dãy tỉ số bằng nhau :

\(\dfrac{a+b+c+d}{a-b+c-d}=\dfrac{a+b-c-d}{a-b-c+d}=\dfrac{\left(a+b+c+d\right)+\left(a+b-c-d\right)}{\left(a-b+c-d\right)+\left(a-b-c+d\right)}=\dfrac{\left(a+b+c+d\right)-\left(a+b-c-d\right)}{\left(a-b+c-d\right)-\left(a-b-c+d\right)}\)

\(\Leftrightarrow\dfrac{2a+2b}{2a-2b}=\dfrac{2c+2d}{2c-2d}\Leftrightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Leftrightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

Áp dụng tính chất thêm một lần nữa , có :

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\dfrac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}\)

\(\Leftrightarrow\dfrac{2a}{2c}=\dfrac{2b}{2d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

5 tháng 8 2017

thanks tui bit lam roihiha

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)=>\(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\)