a^2 + 3b^2=11c^2 + 185d^2,CM:a+3b+11c+d là hợp số
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VE
0
3 tháng 1 2021
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{5a}{3b}=\dfrac{5c}{3d}\)
hay \(\dfrac{5a}{5c}=\dfrac{3b}{3d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Leftrightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
hay \(\dfrac{5a+3n}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)(đpcm)
G
8 tháng 8 2023
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)
a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)\(=\dfrac{\dfrac{a}{k}.b}{\dfrac{c}{k}.d}=\dfrac{ab}{cd}=VT\)
Vậy...
b) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
Suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(bk\right)^2+3\left(bk\right).b}{11\left(bk\right)^2-8b^2}\)\(=\dfrac{7k^2+3k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3\left(dk\right).d}{11\left(dk\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)
Suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
T
8 tháng 8 2023
a) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(ad=bc\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\) => \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
(theo tính chất dãy tỉ số bằng nhau)
=> (đpcm)
b) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)(theo tính chất dãy tỉ số bằng nhau)
=> \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\) (đpcm)
c) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{a^2}{c^2}=\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\) => \(\dfrac{7a^2}{7c^2}=\dfrac{3ab}{3cd}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}\)
=> \(\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\) (theo tính chất dãy tỉ số bằng nhau)
=> \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)(đpcm)
#Ayumu
HN
1
N
13 tháng 10 2019
Sửa chút, chỗ mẫu 11c + 3b thành 11c +3d
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{c}=\frac{b}{d}=\frac{11a}{11c}=\frac{3b}{3d}=\frac{11a+3b}{11c+3d}\\\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{11b}{11d}=\frac{3a-11b}{3c-11d}\end{cases}}\)
\(\Rightarrow\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)
Vậy \(\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)
TN
24 tháng 5 2016
cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k=> a=bk; c=dk
a. Vế trái =\(\frac{5a+3b}{5a-3b}\)=\(\frac{5bk+3b}{5bk-3b}\)=\(\frac{b\left(5k+3\right)}{b\left(5k-3\right)}\)=\(\frac{\left(5k+3\right)}{\left(5k-3\right)}\)(1)
Vế phải =\(\frac{5c+3d}{5c-3d}\)=\(\frac{5dk+3d}{5dk-3d}\)=\(\frac{d\left(5k+3\right)}{d\left(5k-3\right)}\)=\(\frac{\left(5k+3\right)}{\left(5k-3\right)}\)(2)
Từ (1) và (2) ta có\(\frac{5a+3b}{5a-3b}\)=\(\frac{5c+3d}{5c-3d}\)
b. Vế trái=\(\frac{7a^2+3ab}{11a^2-8b^2}\)=\(\frac{7b^2k^2+3b.k.b}{11b^2.k^2-8b^2}\)=\(\frac{b^2.k\left(7k+3\right)}{b^2\left(11k^2-8\right)}\)=\(\frac{k\left(7k+3\right)}{\left(11k^2-8\right)}\)(1)
Vế phải =\(\frac{7c^2+3cd}{11c^2-8d^2}\)=\(\frac{7d^2k^2+3d.k.d}{11d^2.k^2-8d^2}\)=\(\frac{d^2.k\left(7k+3\right)}{d^2\left(11k^2-8\right)}\)=\(\frac{k\left(7k+3\right)}{\left(11k^2-8\right)}\)(2)
Từ (1) và (2) ta có: \(\frac{7a^2+3ab}{11a^2-8b^2}\)=\(\frac{7c^2+3cd}{11c^2-8d^2}\)
5 tháng 4 2017
a) dk: \(\left\{{}\begin{matrix}a,d\ne0\\5a\ne3b\\5c\ne3d\end{matrix}\right.\) \(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5.\dfrac{a}{b}+3}{5\dfrac{a}{b}-3}=\dfrac{5.\dfrac{c}{d}+3}{5\dfrac{c}{d}-3}=\dfrac{\dfrac{5c+3d}{d}}{\dfrac{5c-3d}{d}}=\dfrac{5c+3d}{d}.\dfrac{d}{5c-3d}=\dfrac{5c+3d}{5c-3d}=VP\)
b)
\(\left\{{}\begin{matrix}b,d\ne0\\11a^2\ne8b^2\\11c^2\ne8d^2\end{matrix}\right.\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\right)\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7.\dfrac{a^2}{b^2}+3\dfrac{a}{b}}{11\dfrac{.a^2}{b^2}-8}=\dfrac{7.\dfrac{c^2}{d^2}+3\dfrac{c}{d}}{11\dfrac{.c^2}{d^2}-8}=\dfrac{7c^2+3cd}{11c^2-8d^2}=VP\)
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