cho a/b = c/d . chứng minh (a + 4c ) (2b -3d) = (b +4d) (2 a-3 c) bằng t/c dãy tỉ số bằng nhau
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TT
1
26 tháng 6 2023
`(a+4c)(2b-3d) = 2ab +8bc -3ad - 12cd.`
`(b+4d)(2a-3c) = 2ab + 8ad - 3cb - 12cd`.
Mà do `a/b = c/d => ac = bd => dpcm`.
VQ
1
1 tháng 11 2021
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{2a+3c}{3a+4c}=\dfrac{2bk+3dk}{3bk+4dk}=\dfrac{2b+3d}{3b+4d}\)
20 tháng 12 2016
Giải:
Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,b=ck,c=dk\)
Ta có:
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{bk+ck-dk}{b+c-d}\right)^3=\left[\frac{k\left(b+c-d\right)}{b+c-d}\right]^3=k^3\) (1)
\(\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^2=\left(\frac{2bk+3ck-4dk}{2b+3c-4d}\right)^3=\left[\frac{k\left(2b+3c-4d\right)}{2b+3c-4d}\right]^3=k^3\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^3\) ( đpcm )
15 tháng 8 2023
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(b+d\right)c=\left(a+c\right)d\)
\(\Rightarrow dpcm\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2a}{2b}=\dfrac{c}{d}=\dfrac{2a+c}{2b+d}=\dfrac{2a-c}{2b-d}\)
\(\Rightarrow\left(2b-d\right)\left(2a+c\right)=\left(2a-c\right)\left(2b+d\right)\)
\(\Rightarrow dpcm\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{3a}{3b}=\dfrac{5c}{5d}=\dfrac{3a+5c}{3b+5d}=\dfrac{a-3c}{b-3d}\)
\(\Rightarrow\left(b-3d\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
\(\Rightarrow dpcm\)
15 tháng 8 2023
Đính chính câu c
\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
K
5 tháng 2 2022
đặt a/b =c/d =k
=> a=bm , c=dm
=> 2a+3c/2b+3d =2bm+3bm/ 2b +3d = m.(2d+3d)/2d+3d =m (1)
=> 2a-3c/2d-3d=2bm-3dm /2b -3d =m.(2b-3d)/2b-3d= m (2)
Từ (1) và (2) => 2a+3c/2b+3d =2a-3c/2b-3d
câu 2 tương tự nha
11 tháng 11 2023
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Lời giải:
Ta có:
$\frac{a}{b}=\frac{c}{d}=\frac{4c}{4d}=\frac{a+4c}{b+4d}$ (theo TCDTSBN)
$\frac{a}{b}=\frac{c}{d}=\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a-3c}{2b-3d}$ (theo TCDTSBN)
$\Rightarrow \frac{a+4c}{b+4d}=\frac{2a-3c}{2b-3d}$
$\Rightarrow (a+4c)(2b-3d)=(2a-3c)(b+4d)$ (đpcm)