#)Giải :

b) Vì ba số \(\frac{2006}{2005};\frac{1998}{1997};\frac{82}{81}\)đều có tử số lớn hơn mẫu số ( loại )

    So sánh :

    \(\frac{13}{18}=\frac{13x3}{18x3}=\frac{39}{54};\frac{37}{54}\)

    Vì \(\frac{39}{54}>\frac{37}{54}\Rightarrow\frac{13}{18}>\frac{37}{54}\Rightarrow\frac{37}{54}\)Là phân số bé nhất 

       #~Will~be~Pens~#

#)Giải :

a)     11 x ( y - 6 ) = ( 4 x y ) + 11

        11 x y - 66   = 44 x y + 11

     y x ( 11 + 11 ) = 66 + 44 

     y x 22             = 110

     y                    = 110 : 22

     y                    = 5

        #~Will~be~Pens~#

Trả lời :

a)                    \(11.\left(y-6\right)=\left(4.y\right)+11\)

    \(11.\left(y-6\right)-\left(4.y\right)=11\)

           \(\left(y-6\right)-\left(4.y\right)=11:11\)

           \(\left(y-6\right)-\left(4.y\right)=1\)

    \(y^2-6.y-4.y-6.y=1\)

                              \(y^2-4.y=1\)

                                    \(y-4=1\)

                                             \(y=1+4\)

\(\Rightarrow y=5\)

b) Phân số bé nhất là : \(\frac{13}{18}\)vì  \(\frac{13}{18}< \frac{37}{54}< 1\)và các phân số  \(\frac{2006}{2005},\frac{1998}{1997},\frac{82}{81}>1\).

Hok_Tốt

#Thiên_Hy

Ta có:

\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)

\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)

c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.

d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)

\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)

\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)

theo bài ra ta có:

\(C=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)\)

C = \(\left(\frac{3}{4}-81\right)...\left(\frac{3^6}{9}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)\)

\(C=\left(\frac{3}{4}-81\right)...\left(81-81\right)...\left(\frac{3^{2013}}{2016}-81\right)\)

C = \(\left(\frac{3}{4}-81\right)...0...\left(\frac{3^{2013}}{2016}-81\right)\)

C = 0

vậy C = 0

(3/4 -81 )(3^2/5 -81 )(3^3/6 -81)......(3^2000/2003 -81)

ta viết tiếp dãy số (3/4 -81 )(3^2/5 -81 )(3^3/6 -81)(3^4/7 - 81 ) (3^5/8 -81)(3^6/9 -81).........(3^2000/2003 -81) thì thấy 3^6/9=81 ->3^6/9 -81=0 -> dãy số bằng 0 -> (3/4 -81 )(3^2/5 -81 )(3^3/6 -81)......(3^2000/2003 -81) =0

Minh k hiểu cho lắm. Bạn viết theo công thức toán olm cho sẵn đi cho dễ đọc

\(\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)....\left(\frac{3^{2000}}{2003}-81\right)\)

\(=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^6}{9}-81\right)...\left(\frac{3^{2000}}{2003}-81\right)\)

\(=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)....\left(81-81\right)...\left(\frac{3^{2000}}{2003}-81\right)\)

\(=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)....0....\left(\frac{3^{2000}}{2003}-81\right)\)

\(=0\)

Ta có \(81.\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{3}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right)\)

\(=81.\left(\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right)\)

\(=81.\left(\frac{12}{4}:\frac{5}{6}\right)\)

\(=81.\frac{18}{5}\)

\(=291,6\)

\(81\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{159}+\frac{6}{91}}\right)\)

\(=81\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{2}{169}+\frac{1}{91}\right)}\right)\)

\(=81\left(3\div\frac{5}{6}\right)\)

\(=81.\frac{18}{5}\)

\(=\frac{1458}{5}\)

81.[12(1-1/7-1/289-1/85)/4(1-1/7-1/289-1/85):5(1+1/13+1/169+1/91)/6(1/13+1/169+1/91)

=81.[12/4:5/6]=81.[12/4.6/5]=81.18/5=1458/5

6/5 > 5/6

8/9 > 21/81

9/63 < 90/7

8/81 < 10/9

Tk mk nha

so sánh với 1

rút gọn hoặc quy đồng số còn lại để so sánh tử

so sánh với 1

cũng so sánh với 1

bạn đùa hay hỏi thật vậy

\(\frac{6}{5}>\frac{5}{6}\)

\(\frac{8}{9}>\frac{21}{28}\)

\(\frac{9}{63}< \)\(\frac{90}{7}\)

\(\frac{8}{81}< \frac{10}{9}\)

Trong biểu thức trên có một thừa số là : \(\frac{3^6}{9}-81\)

Ta có : \(\frac{3^6}{9}-81=\frac{3^6}{3^2}-81=3^4-81=81-81=0\)

Do đó giá trị biểu thức trên bằng 0

sai đề 100%

Đề bài không sai đâu bạn. Đó là bài của vòng 17 Toán Violympic năm 2016