yutyugubhujyikiu

A= (-4/5+4/3)+(-5/4+14/5)-7/3

= 8/15+31/20-7/3

= 25/12-7/3

= -1/4

B= 8/3.2/5.3/8.10.19/92

= 16/15.3/8.10.19/92

= 2/5.10.19/92

= 4.19/92

= 19/23

C= \(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+1\(\frac{5}{7}\)

=\(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)

= \(\frac{-10}{77}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)

= \(\frac{-10}{77}\)+\(\frac{-45}{98}\)+\(\frac{12}{7}\)

= \(\frac{-635}{1078}\)+\(\frac{12}{7}\)

= \(\frac{1213}{1078}\)

thanks nhìu

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

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:

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

Mình làm như thế này nek

\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

a) \(\frac{14}{21}+1-\left|\frac{1}{3}-1\right|\)

\(=\frac{2}{3}+1-\frac{2}{3}\)

\(=1+\left(\frac{2}{3}-\frac{1}{3}\right)\)

\(=1\)

b) \(\frac{1}{3}-\left|\frac{-1}{4}+\frac{5}{6}\right|-\left|\frac{-7}{12}\right|\)

\(=\frac{1}{3}-\frac{7}{12}-\frac{7}{12}\)

\(=-\frac{5}{6}\)

Các câu khác làm tương tự thôi :))

\(\frac{1}{3}-|\frac{-1}{4}+\frac{5}{6}|-|\frac{-7}{12}|\)

\(=\frac{2}{3}+1-\frac{2}{3}\)

\(=1+(\frac{2}{3}-\frac{2}{3})\)

\(=1\)

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)

a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)

\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} =  \frac{{ -29}}{{35}}\end{array}\)

b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)

\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)

c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)

\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)

d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)

\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ =  (- 2).45,3 =  - 90,6\end{array}\)

a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)

b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)

\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)

c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)

d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)

\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)

\(A=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)

\(A=\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}-\frac{7}{3}\right)+\frac{-5}{4}\)

\(A=2+\left(-1\right)+\frac{-5}{4}\)

\(A=\frac{-1}{4}\)

A=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)

  =\(\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}+\frac{-7}{3}\)=\(\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}+\frac{-7}{3}\right)+\frac{-5}{4}\)

  =\(\frac{10}{5}+\frac{-3}{3}+\frac{-5}{4}\)=\(2-1+\frac{-5}{4}\)=\(1+\frac{-5}{4}\)=\(\frac{4}{4}+\frac{-5}{4}\)=\(\frac{4-5}{4}\)=\(\frac{-1}{4}\)

\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)

\(=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)

\(=\frac{-4}{5}+\frac{14}{5}+\frac{4}{3}-\frac{7}{3}+\frac{14}{5}\)

\(=-2-1+\frac{14}{5}\)

\(=\frac{14}{5}-\frac{15}{5}\)

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