ủa! Nên xem lại đề pn ạ! Nếu mak tim x thì kết quả đâu mak nếu lak tính thì phải loại chữ số x

King Of Fairy In The Sky x là nhân ông bạn à quên kiến thức à

x là dấu nhân, còn x kia là chữ x 

khó thế thế mà bào toán lớp 1

ĐÁP SỐ LÀ : 1 

x=1 nha!!!

{ [ 27 * 11 - ( x + 12 ) * 8 ] * 3 - 79 } : 5 = 100

               [ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 100 * 5

               [ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 500

                       [ 297 - ( x + 12 ) * 8 ] * 3 = 500 + 79

                       [ 297 - ( x + 12 ) * 8 ] * 3 = 579

                                297 - ( x + 12 ) * 8 = 579 : 3

                                297 - ( x + 12 ) * 8 = 193

                                          ( x + 12 ) * 8 = 297 - 193

                                          ( x + 12 ) * 8 = 104

                                                    x + 12 = 104 : 8

                                                    x + 12 = 13

                                                            x = 13 - 12

                                                            x = 1.^^             

(297-(x+12)*8)*3-79=297-79-(x+12)*8*3

                            =218-(x+12)*24

                           =218-24x+288

                           =218-288+24x

                           =-70+24x

Kết quả là

  x = 1

       Đs...

x = 1 nha 

\(x=1\)

Kết quả là

 X = 1

   Đs...

\(\left\{\left[27\times11-\left(x+12\right)\times8\right]\times3-79\right\}\div5=100\)

\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=100\times5\)

\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=500\)

\(\left[27\times11-\left(x+12\right)\times8\right]\times3=500+79\)

\(\left[27\times11-\left(x+12\right)\times8\right]\times3=579\)

\(27\times11-\left(x+12\right)\times8=579:3\)

\(27\times11-\left(x+12\right)\times8=193\)

\(297-\left(x+12\right)\times8=193\)

\(\left(x+12\right)\times8=297-193\)

\(\left(x+12\right)\times8=104\)

\(x+12=104:8\)

\(x+12=13\)

\(x=13-12\)

\(x=1\)

\(k\)\(dung\)\(cho\)\(minh\)\(nhe!\)

x = 1 nha 

kết quả là

x=1

    ĐS///

a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79

= 1/10 - 1/79

= máy tính ok

mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha

a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)

Bài làm:

a) \(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}\)

\(=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}\)

\(=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{135}\right)\)

\(=4.\frac{128}{945}=\frac{512}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}\)

\(=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{506}\right)\)

\(=4.\frac{249}{2024}=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}\)

\(=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{394}\right)\)

\(=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}\)

\(=\frac{4}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)\)

\(=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Phải sửa \(\frac{1}{802}\)  thành \(\frac{1}{820}\) nhé

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)

\(=1+\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{1}{41.20}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}\right)\)

\(=2\left(1-\frac{1}{41}\right)\)

\(=2.\frac{40}{41}=\frac{80}{41}\)