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.

Cảm ơn

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:

\(\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\)

\(ab+ad+2cb+2cd=ab+2ad+cb+2cd\)

\(cb=ad\Rightarrow\frac{a}{b}=\frac{c}{d}\)

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 )

\(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}\Rightarrow\frac{a+2c}{b+2d}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a+c}{b+d}\)

\(\Rightarrow\frac{a+2c}{b+2d}=\frac{a+c}{b+d}\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\)

a)  đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow a=b.k;c=d.k\)

          \(\frac{3a+2c}{3b+2d}=\frac{3b.k+2.d.k}{3b+2d}=\frac{k\left(3b+2d\right)}{3b+2d}=k\)

    b)          bó tay

mk trả lời như thế này có đúng không các bạn góp ý nhé

vì a/b=c/d = \(\frac{a+c}{b+d}\left(1\right)\)

ta lại có:

a/b=c/d=\(\frac{a+2c}{2d}=\frac{a+2c}{b+2d}\left(2\right)\)

từ 1 và 2 ta có:

=>(a+2c).(b+d)=(a+c).(b+2d)

đúng rồi đó bạn
k cho mình ?