\(A=\frac{1}{3}.\frac{-9}{10}.\frac{-6}{13}.\frac{-13}{36}=\frac{-3}{10}.\frac{-1}{6}=\frac{1}{20}\)

\(B=\frac{4}{19}\left(\frac{-5}{12}+\frac{-7}{12}\right)-\frac{40}{57}=\frac{-4}{19}-\frac{40}{57}=\frac{-52}{57}\)

2 câu còn lại tự làm

\(A=\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)

\(A=\frac{1}{1}.\frac{-2}{13}.\frac{-9}{10}.\frac{-13}{36}\)

\(A=\frac{-2}{13}.\frac{-9}{10}.\frac{-13}{36}\)

\(A=\frac{-1}{13}.\frac{-9}{5}.\frac{-13}{36}\)

\(A=\frac{-1}{13}.\frac{-1}{5}.\frac{-13}{4}\)

\(A=\frac{-13}{260}=\frac{-1}{20}\)

Câu 1a:

-3/8 + 5/-12

= -9/24 - 10/24

= - 19/24

Câu 1b:

3/14 + - 2/7

= 3/14 - 4/14

= - 1/14

Câu 1c:

3/5 - -7/10 - 13/-20

= 12/20 + 14/20 + 13/20

= 26/20 + 13/20

= 39/20

= 39/20

Câu 1d:

1/4 - 5/3.3/15

= 1/4 - 1/3

= 3/12 - 4/12

= - 1/12

Câu 1e:

-3/5 + 2/7 + 2/-5 + 3/5 + 5/7

= (-3/5 + 3/5 - 2/5) + (2/7 + 5/7)

= (0 - 2/5) + 1

= - 2/5 + 1

= - 2/5 + 5/5

= 3/5

0,7.2\(\frac23\).20.0,375.\(\frac{5}{28}\)

= \(\frac{7}{10}\) .\(\frac83\).20.\(\frac38\).\(\frac{5}{28}\)

= (\(\frac{7}{10}\).\(\frac{5}{28}\).20).(\(\frac83\).\(\frac38\))

= (\(\frac18.20\)). 1

= \(\frac52\)

B = (9,75 .21\(\frac37\) + \(\frac{39}{4}\).18\(\frac47\))

B = (\(\frac{39}{4}\).\(\frac{150}{7}\frac{}{7}\)+ \(\frac{39}{4}\).\(\frac{130}{7}\))

B = \(\frac{39}{4}\).(\(\frac{150}{7}\) + \(\frac{130}{7}\))

B = \(\frac{39}{4}\).\(\frac{280}{7}\)

B = 390

1. a) \(\frac{-2}{7}+\frac{15}{23}+\frac{\left(-15\right)}{17}+\frac{4}{19}+\frac{8}{23}\)

    \(=\left(\frac{-2}{7}+\frac{-5}{7}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)

    \(=\left(-1\right)+1+\frac{4}{19}\)

    \(=0+\frac{4}{19}=\frac{4}{19}\)

b) \(\frac{7}{19}\cdot\frac{8}{11}+\frac{7}{19}\cdot\frac{3}{11}+\frac{12}{19}\)

   \(=\frac{7}{19}\cdot\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)

   \(=\frac{7}{19}\cdot1+\frac{12}{19}\)

   \(=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1\)

2. a) \(\frac{1}{3}+\frac{\left(-2\right)}{16}-\frac{7}{14}\)

\(=\frac{5}{24}-\frac{1}{2}\)

\(=-\frac{7}{24}\)

b) \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)

\(=\left(11-2+5\right)+\frac{3}{13}-\frac{4}{7}+\frac{3}{13}\)

\(=14+\left(-\frac{10}{91}\right)\)

\(=-14\frac{10}{91}\)

c) \(0,7\cdot2\frac{2}{3}\cdot20\cdot0,375\cdot\frac{5}{28}\)

\(=\frac{7}{10}\cdot\frac{8}{3}\cdot20\cdot\frac{3}{8}\cdot\frac{5}{28}\)

\(=\left(\frac{7}{10}\cdot\frac{5}{28}\right)\cdot\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot20\)

\(=\frac{1}{8}\cdot1\cdot20\)

\(=\frac{20}{8}=\frac{5}{2}\)

d) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{1}{7}-\frac{8}{9}\)

\(=1-\frac{8}{9}\)

\(=\frac{1}{9}\)

~Học tốt~

Bài 1: Tìm \( x \)

\[
x - \frac{25\%}{100}x = \frac{1}{2}
\]

Để giải phương trình này, trước hết chúng ta phải chuyển đổi phần trăm thành dạng thập phân:

\[
\frac{25\%}{100} = 0.25
\]

Phương trình ban đầu trở thành:

\[
x - 0.25x = \frac{1}{2}
\]

Tổng hợp các hạng tử giống nhau:

\[
1x - 0.25x = \frac{1}{2}
\]
\[
0.75x = \frac{1}{2}
\]

Giải phương trình ta được:

\[
x = \frac{\frac{1}{2}}{0.75} = \frac{2}{3}
\]

Vậy, \( x = \frac{2}{3} \)

Bài 2: Tính hợp lý

a) \[
\frac{5}{-4} + \frac{3}{4} + \frac{4}{-5} + \frac{14}{5} - \frac{7}{3}
\]

Chúng ta cần tìm một mẫu số chung cho tất cả các phân số. Mẫu số chung nhỏ nhất là 60.

\[
= \frac{75}{-60} + \frac{45}{60} + \frac{-48}{60} + \frac{168}{60} - \frac{140}{60}
\]
\[
= \frac{75 + 45 - 48 + 168 - 140}{60}
\]
\[
= \frac{100}{60} = \frac{5}{3}
\]

b) \[
\frac{8}{3} \times \frac{2}{5} \times \frac{3}{10} \times \frac{10}{92} \times \frac{19}{92}
\]

Tích của các phân số là:

\[
= \frac{8 \times 2 \times 3 \times 10 \times 19}{3 \times 5 \times 10 \times 92 \times 92}
\]
\[
= \frac{9120}{4131600} = \frac{57}{25825}
\]

c) \[
\frac{5}{7} \times \frac{2}{11} + \frac{5}{7} \times \frac{9}{14} + \frac{1}{5}
\]

Tích của các phân số là:

\[
= \frac{10}{77} + \frac{45}{98} + \frac{1}{5}
\]
\[
= \frac{980}{7546} + \frac{3485}{7546} + \frac{15092}{75460}
\]
\[
= \frac{2507}{7546}
\]

a

\(5\frac{4}{7}:x+=13\)

\(\frac{39}{7}:x=13\)

\(x=\frac{39}{7}:13\)

\(x=\frac{3}{7}\)

\(\frac{4}{7}x=\frac{9}{8}-0,125\)

\(\frac{4}{7}x=1\)

\(x=1:\frac{4}{7}\)

\(x=\frac{7}{4}=1\frac{3}{4}\)

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}