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Câu 8b:

\(4\dfrac{209}{245}:\left(\dfrac{3}{5}+\dfrac{4}{7}\right)+4\dfrac{209}{245}:\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\)

\(=4\dfrac{209}{245}:\dfrac{41}{35}+4\dfrac{209}{245}:\dfrac{29}{35}\)

\(=4\dfrac{209}{245}\cdot\dfrac{35}{41}+4\dfrac{209}{245}\cdot\dfrac{35}{29}\)

\(=\left(4+\dfrac{209}{245}\right)\left(\dfrac{35}{41}+\dfrac{35}{29}\right)\)

\(=\dfrac{1189}{245}\cdot35\left(\dfrac{1}{41}+\dfrac{1}{29}\right)\)

\(=\dfrac{1189}{7}\cdot\dfrac{70}{1189}=\dfrac{70}{7}=10\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)

\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}\)

Do đó: \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

(a+b+c+d)(a+d-b-c)=(a-b+c-d)(a+b-c-d)

=>(a+d)^2-(b+c)^2=(a-d)^2-(b-c)^2

=>(a+d)^2-(a-d)^2=(b+c)^2-(b-c)^2

=>(a+d-a+d)(a+d+a-d)=(b+c+b-c)(b+c-b+c)

=>4ad=4bc

=>ad=bc

=>a/c=b/d

28 tháng 6 2021

- OBC = OAD ( g - c - g )

- DEB = CEA ( g - c - g )

 

 

\(x:\left[\dfrac{8}{5}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{5}\right]=\dfrac{15}{7}+\dfrac{6}{5}\left[\left(2\dfrac{1}{7}\right)^2-\dfrac{50}{49}\right]\)

\(\Leftrightarrow x:\left[\dfrac{32}{45}-\dfrac{18}{45}\right]=\dfrac{15}{7}+\dfrac{6}{5}\cdot\left(\dfrac{225}{49}-\dfrac{50}{49}\right)\)

\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{15}{7}+\dfrac{6}{5}\cdot\dfrac{25}{7}\)

\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{45}{7}\)

\(\Leftrightarrow x=2\)