Bài 1 : Ta có:

\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)

= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)

= \(\frac{7}{9}\)

Bài 2 :

 \(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)

=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)

=> 50x = 10

=> x = 10 : 50

=> x = 1/5

Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3

                                         <=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy 

\(-\dfrac{3}{4}.31\dfrac{11}{23}-0,75.8\dfrac{12}{23}\)

\(=-\dfrac{3}{4}.31\dfrac{11}{23}-\dfrac{3}{4}.8\dfrac{12}{23}\)

\(=-\dfrac{3}{4}.\left(31+\dfrac{11}{23}+8+\dfrac{12}{12}\right)\)

\(=-\dfrac{3}{4}.\left(31+8+1\right)\)

\(=-\dfrac{3}{4}.40\)

\(=-3.10\)

\(=-30\)

- \(\dfrac{3}{4}\).31\(\dfrac{11}{23}\) - 0,75.8\(\dfrac{12}{23}\)

= - \(\dfrac{3}{4}\).\(\dfrac{724}{23}\) - \(\dfrac{3}{4}\)\(\dfrac{196}{23}\)

=   - \(\dfrac{3}{4.23}.\left(724+196\right)\)

= - \(\dfrac{3}{92}\) . 920

= - 30 

a, 1 - 7x = 3x - 4

=> -7x - 3x = - 4 - 1

=> - 10x = - 5

=> x = 1/2

vậy_

b, đặt  \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

mk chỉ bt lm mấy phần hui à!

d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)

\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)

e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)

mk xl,mk ko bt lm các phần còn lại!

a) Ta có: \(C=-\left|x+2\right|\le0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|x+2\right|=0\Rightarrow x=-2\)

Vậy Max(C) = 0 khi x = -2

b) Ta có: \(D=1-\left|2x-3\right|\le1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|2x-3\right|=0\Rightarrow x=\frac{3}{2}\)

Vậy Max(D) = 1 khi x = 3/2

d) \(D=-\left|x+\frac{5}{2}\right|\le0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|x+\frac{5}{2}\right|=0\Rightarrow x=-\frac{5}{2}\)

Vậy Max(D) = 0 khi x = -5/2

e) \(P=4-\left|5x-3\right|-\left|3y+12\right|\le4\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|5x-3\right|=0\\\left|3y+12\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{5}\\y=-4\end{cases}}\)

Vậy Max(P) = 4 khi \(\hept{\begin{cases}x=\frac{3}{5}\\y=-4\end{cases}}\)

\(-\dfrac{3}{4}.31\dfrac{11}{23}-0,75.8\dfrac{11}{23}\)

\(=\dfrac{3}{4}.\left(-31\dfrac{11}{23}\right)-\dfrac{3}{4}.8\dfrac{11}{23}\)

\(=\dfrac{3}{4}.\left(-31\dfrac{11}{23}-8\dfrac{11}{23}\right)\)

\(=\dfrac{3}{4}.\left(-39\right)\)

\(=-\dfrac{117}{4}\)

giúp mk vs

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

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

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

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> \(4\left(x+1\right)=3\left(x-2\right)\)

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)