Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)
Thay (1) vào đề:
\(VT=\left(2a+3c\right)\left(b+d\right)=\left(2bk+3dk\right)\left(b+d\right)=2b^2k+3bdk+2bdk+3d^2k=3d^2k+2b^2k+5bdk\)
\(VP=\left(bk+dk\right)\left(2b+3d\right)=2b^2k+2bdk+3bdk+3d^2k=3d^2k+2b^2k+5bdk\)
Khi đó: \(VT=VP\)
\(\Leftrightarrow\left(2a+3c\right)\left(b+d\right)=\left(a+c\right)\left(2b+3d\right)\rightarrowđpcm.\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\left(2a+3c\right)\left(b+d\right)=\left(2bk+3dk\right)\left(b+d\right)=2b^2k+2bkd+3bkd+3d^2k\)
\(=2b^2k+5bkd+3d^2k\)(1)
\(\left(a+c\right)\left(2b+3d\right)=\left(bk+dk\right)\left(2b+3d\right)=2b^2k+3bkd+2bkd+3d^2k\)
\(=2b^2k+5bkd+3d^2k\)(2)
Từ (1) và (2) suy ra:
\(\left(2a+3c\right).\left(b+d\right)=\left(a+c\right)\left(2b+3d\right)\)(đpcm)
Chúc bạn học tốt!!!
Theo đề bài ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> a = bk
=> c = dk
Ta có :
(2a + 3c)(b + d) = (2bk + 3dk)(b + d) = k(2b + 3d)(b + d)(1)
(a + c)(2b + 3d) = (bk + dk)(2b + 3d) = k(2d + 3d)(b + d)(2)
Từ (1) và (2)
=> (2a + 3c)(b + d) = (a + c)(2b + 3d)(đpcm)
tích nha .....
Ta có:
\((2a+3c)(b+d)=2a(b+d)+3c(b+d)\)
\(=2ab+2ad+3cb+3cd\)
\((a+c)(2b+3d)=a(2b+3d)+c(2b+3d)\)
\(=2ab+3ad+2bc+3dc\)
đpcm:\(2ab+2ad+3cb+3cd=\)\(2ab+3ad\)\(+2cb+3dc\)
Rút gọn,ta có đpcm=
\(2ad+\)\(3cb\)\(+3cd=\)\(3ad+2cb+3dc\)
\(2ad+3cb=3ad+2cb\)
\(cb=ad\)
Mà,theo đề bài :
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\left(đpcm\right)\)
