B1      :

\(\frac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}\) . x = 0,(2)

=\(\frac{0,5}{1,5}\).x=0,(2)

x=0,(2):\(\frac{0,5}{1,5}\)

x=0,(6)=\(\frac{2}{3}\)

b2:

 [12,(1) - 2,3(6)] : 4,(21)

=9,7(4):4,(21)

=\(\frac{9,7\left(4\right)}{4,\left(21\right)}\)

di ban

\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

\(A=\left(\frac{\frac{3}{2}+1-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)

\(A=\left(\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{-3.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}\right).\frac{401}{378}+115\)

\(A=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)

\(A=0.\frac{401}{378}+115=115\)

A = \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

= \(\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3.125}{100}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5.125}{100}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

= \(\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\right):\frac{1890}{2005}+115\)

= \(\left(\frac{3}{5}+-\frac{3}{5}\right):\frac{1890}{2005}+115\)

= 115

\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

\(=\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)

\(=\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\right)}\right):\frac{378}{401}+115\)

\(=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)

\(=0.\frac{401}{378}+115\)

\(=115\)

Vậy A = 115

bài 1)

70:2=35(m)

Gọi a và b lần lượt là chiều rộng và chiều dài của miếng đất

Từ b/a = 4 /3 = > 3/a = 4 /b

= > 3/ a = 4/ b = 3 + 4/ a + b = 7/ 35 = 5 /3 a = 5

= > a = 3.5 = 15/ 4 b = 5

= > b = 5.4 = 20

Vậy diện tích miếng đất đó là:

15.20=300(m2)

2) Bài 138 (Sách bài tập - tập 1 - trang 33)

 bài 2 cậu vào cái ý là có 

A = -2 nhé , 

gợi ý các bạn chọn (k) đúng cho mình.

áp dung \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)

\(\dfrac{7}{9}+\left(\dfrac{2}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{9}+\left(\dfrac{8}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{9}+\dfrac{1}{4}\)
\(=\dfrac{28}{36}+\dfrac{9}{36}=\dfrac{37}{36}\)
\(-------------\)
\(\dfrac{5}{8}+\dfrac{1}{4}\times\dfrac{2}{3}\)
\(=\dfrac{5}{8}+\dfrac{1}{6}\)
\(=\dfrac{15}{24}+\dfrac{4}{24}=\dfrac{19}{24}\)

\(\dfrac{7}{9}+\left(\dfrac{2}{3}-\dfrac{5}{12}\right)=\dfrac{7}{9}+\dfrac{1}{4}=\dfrac{28}{36}+\dfrac{9}{36}=\dfrac{37}{36}\)

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\(\dfrac{5}{8}+\dfrac{1}{4}\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{1}{6}=\dfrac{15}{24}+\dfrac{4}{24}=\dfrac{19}{24}\)