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từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\)ad = bc \(\Rightarrow\)ad + 2bc = bc + 2ad
\(\Rightarrow\)ab + ad + 2bc + 2cd = ab + 2ad + bc + 2cd
\(\Rightarrow\)a ( b + d ) + 2c ( b + d ) = a ( b + 2d ) + c ( b + 2d )
\(\Rightarrow\)( a + 2c ) ( b + d ) = ( a + c ) ( b + 2d )
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(a+2c\right)\left(b+d\right)=\left(bk+2dk\right)\left(b+d\right)=k\left(b+2d\right)\left(b+d\right)\)
\(\left(a+c\right)\left(b+2d\right)=\left(bk+dk\right)\left(b+2d\right)=k\left(b+d\right)\left(b+2d\right)\)
Do đó: VT=VP
Lời giải:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow d=bk; c=dk\). Thay vào biểu thức ta có:
\((a+2c)(b+d)=(bk+2dk)(b+d)=k(b+2d)(b+d)(*)\)
\((a+c)(b+2d)=(bk+dk)(b+2d)=k(b+d)(b+2d)(**)\)
Từ \((*); (**)\Rightarrow (a+2c)(b+d)=(a+c)(b+2d)\)
Ta có đpcm.
Có: a/b=c/d⇒a/b=c/d=2c/2d
=a+c/b+d=a+2c/b+2d
⇒(a+C)(a+2c)=(b+d)(b+2d)
ta có: \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\frac{a}{b}=k\Rightarrow a=bk\)
\(\frac{c}{d}=k\Rightarrow c=dk\)
thay vào \(\left(a+2c\right).\left(b+d\right)=\left(bk+2dk\right).\left(b+d\right)=k.\left(b+2d\right).\left(b+d\right)\)
\(\left(a+c\right).\left(b+2d\right)=\left(bk+dk\right).\left(b+2d\right)=k.\left(b+d\right).\left(b+2d\right)\)
\(\Rightarrow\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\left(=k.\left(b+2d\right).\left(b+d\right)\right)\)( đ p c m)
CHÚC BN HỌC TỐT!!!!!!!!
Ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
mà \(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}\)
\(\Rightarrow\frac{a+c}{b+d}=\frac{a+2c}{b+2d}\)\(\Rightarrow\left(a+c\right)\left(b+2d\right)=\left(b+d\right)\left(a+2c\right)\)( đpcm )