a) B = \(\frac{1}{2}.\frac{4}{3}.20.\frac{7}{35}.0,75\)

B = \(\frac{1.1.2.1.1}{1.1.1.1.1}\)

B= 2

b) \(\frac{1}{3}x=16\frac{1}{4}-13\frac{1}{4}\)

= > \(\frac{1}{3}x=3\)

= > x = 9

c)\(\frac{2}{3}x=\frac{-1}{4}-\frac{1}{3}\)

=> \(\frac{2}{3}x=-\frac{7}{12}\)

=> x = \(\frac{-7}{12}:\frac{2}{3}\)

=> x = \(\frac{-7}{8}\)

\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)

\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)

\(< =>\frac{-15+12-24}{63}\)

\(< =>\frac{-3}{7}\)

\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{7}\)

\(< =>\frac{29}{35}\)

\(bai2:\)

\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)

\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)

\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)

\(< =>x=\frac{3}{5}:\frac{-3}{4}\)

\(< =>x=\frac{-4}{5}\)

\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)

\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)

Bài 1:

a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)                                 b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

 \(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\)                                       \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\) 

  \(=\frac{3}{7}.\frac{-9}{9}\)                                                                  \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

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

                                                                                               \(=\frac{29}{35}\)

Bài 2:

a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)                                               b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

  \(\frac{-3}{4}x\)           \(=\frac{1}{5}+\frac{4}{10}\)                                     \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\frac{-3}{4}x\)             \(=\frac{3}{5}\)                                            \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)     

         \(x\)              \(=\frac{3}{5}:\frac{-3}{4}\)                                        \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)                                         

         \(x\)              \(=\frac{4}{-5}\)                                                   \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)

                                                                                                             \(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\) 

                                                                                                                          \(x.\frac{10}{3}=\frac{13}{12}\) 

                                                                                                                                    \(x=\frac{13}{12}:\frac{10}{3}\) 

                                                                                                                                     \(x=\frac{13}{40}\)                             

Bài 3:

a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)

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

2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)

3A = \(1-\frac{1}{2^6}\)

=> 3A < 1 

=> A < \(\frac{1}{3}\)(đpcm)

b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)

4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)       (1)

Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)

4B = \(3-\frac{1}{3^{99}}\)

=> 4B < 3

=> B < \(\frac{3}{4}\)   (2)

Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)

bài 1:

5n+7 chia hết cho 3n+2

=> [3(5n+7) - 5(3n + 2)] chia hết cho 3n+2

=> (15n + 21 - 15n - 10) chia hết cho 3n+2

=> 11 chia hết cho 3n + 2

=> 3n + 2 thuộc Ư(11) = {1;-1;11;-11}

Ta có bảng:

Vậy n = {-1;3}

Bài 2:

1, chữ số tận cùng

a, Xét 7¹⁹⁹⁹

Ta có: 7¹⁹⁹⁹ = 7¹⁹⁹⁶.7³ = (7⁴)⁴⁹⁹.343 = (...1)⁴⁹⁹.343 = (....1).343 = ....3 (1)

Vậy số 57¹⁹⁹⁹ có tận cùng là 3

b, Xét 3¹⁹⁹⁹

Ta có: 3¹⁹⁹⁹ = 3¹⁹⁹⁶.3³ = (3⁴)⁴⁹⁹.27 = (...1)⁴⁹⁹.27 = (...1) . 27 = ....7  (2)

Vậy số 93¹⁹⁹⁹ có chữ số tận cùng là 7

2, 

Từ (1) và (2) suy ra A = 999993¹⁹⁹⁹ + 555557¹⁹⁹⁹ = ...7 + ...3 = ....0

Vì A có chữ số tận cùng là 0 nên A chia hết cho 5. 

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

1 

a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)

= \(3\cdot25:\frac{5}{4}\)

= \(3\cdot\left(25:\frac{5}{4}\right)\)

=\(3\cdot20\)

=60

b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)

=\(\frac{3}{7}\cdot\left(-7\right)\)

=\(-3\)

c) = 

\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)

Ta xét 2 trường hợp 

\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)

tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian 

Bài 1: Tìm x, biết:

\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}}\)

Dùng máy tính

Nếu ko có máy tính thì sao?

Tính nháp

a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)

\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)

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

\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)

b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)

\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)

\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)

\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)

c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)

\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)

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

Bài 2  Bạn tự làm nhé

1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)

\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)

\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)

\(=\frac{67}{4}\)

b,Các phép tính khác làm tương tự

Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ

c,tương tự

2.

a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)

\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)

\(\frac{7}{12}\div x=\frac{-77}{20}\)

Đến đây dễ bạn tự làm

b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)

\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)

\(\frac{14}{5}x+50=-34\)

\(\frac{14}{5}x=-84\)

Tự làm tiếp

c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)