Bài 9: Số thập phân hữu hạn. Số thập phân vô hạn tuần hoàn

phía sau quyển này có đáp án. mở ra xem mà tham khảo

$\frac{2}{3}x-1\frac{2}{5}=\frac{1}{2}x+0,8\left(3\right)$

$\frac{2}{3}x-1\frac{2}{5}=\frac{1}{2}x+\mathrm{\backslash (\backslash dfrac\{5\}\{6\}\backslash )}$

$\mathrm{=>\; \backslash (\backslash dfrac\{2\}\{3\}x-\backslash dfrac\{1\}\{2\}x=\backslash dfrac\{5\}\{6\}+\backslash dfrac\{7\}\{5\}\backslash )}$

=> \(\dfrac{1}{6}x=\dfrac{63}{30}\)

=> \(x=\dfrac{63}{30}:\dfrac{1}{6}\)

=> \(x=\dfrac{67}{5}\)

Vậy : x= 67 / 5

\(\dfrac{2}{3}x-1\dfrac{2}{5}=\dfrac{1}{2}x+0,8\left(3\right)\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{2}x=0,8\left(3\right)+1\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{67}{30}\)

\(\Leftrightarrow x=\dfrac{67}{5}\)

Vậy \(x=\dfrac{67}{5}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ⇒ a = b.k ; c = d.k

\(\dfrac{3a-2c}{5a+4c}=\dfrac{3.b.k-2.d.k}{5.b.k+4.d.k}=\dfrac{k\left(3.b-2.d\right)}{k\left(5b+4d\right)}=\dfrac{3b-2d}{5b+4d}\)

\(\dfrac{3b-2d}{5b+4d}=\dfrac{3b-2d}{5b+4d}\Rightarrow\dfrac{3a-2c}{5a+4c}=\dfrac{3b-2d}{5b+4d}\)

Giải:

a) \(2,5\left(7\right)=\dfrac{116}{45}\)

b) \(0,1\left(35\right)=\dfrac{67}{495}\)

c) \(-3,\left(16\right)=-\dfrac{313}{99}\)

d) \(-4,2\left(018\right)=-\dfrac{2332}{555}\)

Chúc bạn học tốt!

a)1,3(12)=\(\dfrac{1}{10}.\left(13+12.\dfrac{1}{99}\right)\)=\(\dfrac{443}{330}\)

b)\(3,3\left(21\right)=\dfrac{1}{10}\left(33+21.\dfrac{1}{99}\right)=\dfrac{548}{165}\)

c)\(5,2\left(24\right)=\dfrac{1}{10}\left(52+24.\dfrac{1}{99}\right)=\dfrac{862}{165}\)

d)\(3,32\left(6\right)=\dfrac{1}{100}\left(332+6.\dfrac{1}{9}\right)=\dfrac{499}{150}\)

A, 1,3(12) = \(\dfrac{433}{330}\)

B, 3,3(21) = \(\dfrac{548}{165}\)

C,5,2(24) = \(\dfrac{862}{165}\)

D,3,32(6) = \(\dfrac{499}{150}\)

A, 1,3(12)=\(\dfrac{433}{330}\)

B, 3,3(21)= \(\dfrac{548}{165}\)

C, 5,2(24)= \(\dfrac{862}{165}\)

D, 3,32(6)= \(\dfrac{499}{150}\)

a) 0,1234567

b) 10,2345

c) 10,234

2261/495

2261/495