(13.15):(13/2-x)=39

195:(13/2-x)=39

13/2-x=195:39

13/2-x=5

x=13/2-5

x=3/2

vậy x=3/2

Ta có: \(\left(7.13+8.13\right):\left(\frac{13}{2-x}\right)=39\)

         \(\Leftrightarrow\)\(\frac{\left(91+104\right).\left(2-x\right)}{13}=\)\(39\)

        \(\Leftrightarrow\) \(\frac{195.\left(2-x\right)}{13}=39\)

        \(\Leftrightarrow\)\(390-195x=39.13\)

        \(\Leftrightarrow\)\(390-195x=507\)

         \(\Leftrightarrow\)\(-195x=117\)

          \(\Leftrightarrow\)x=\(\frac{-117}{195}=\frac{-39}{65}\)

\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)

\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)

\(=17.\frac{1}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)

\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(=\frac{17}{3}.\frac{6}{49}\)

\(=\frac{34}{49}\)

\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)

\(=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)

\(=13.\frac{1}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)

\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(=\frac{13}{3}.\frac{6}{49}\)

\(=\frac{26}{49}\)

Câu C sai đề rồi , phải là như thế này :

\(C=\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)

\(=\frac{4}{5}.\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)

\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)

\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{258}\right)\)

\(=\frac{4}{5}.\frac{125}{1032}\)

\(=\frac{25}{258}\)

a) (10-x)²:3+3=-9

    (10-x)²:3=-9-3

    (10-x)²:3=-11

    (10-x)²=-11.3

    (10-x)²=-33

 Vô lí vì -33 ko chuyển đc mũ 2 

Vậy x e 0

b) |x-6|.7-4=31

    |x-6|=(31+4):7

    |x-6|=5

\(\Rightarrow\orbr{\begin{cases}x-6=5\Rightarrow5+6=11\\x-6=-5\Rightarrow x=-5+6=1\end{cases}}\)

Vậy.....................................................

a) \(\frac{\left(10-x\right)^2}{3}+3=-9\)

\(\Leftrightarrow\frac{\left(10-x\right)^2}{3}=-12\)

\(\Leftrightarrow\left(10-x\right)^2=-36\)( vô nghiệm vì (10-x)² \(\ge\)0 với mọi x thuộc Z)

b) |x-6|.7-(-2)²=-8+39

<=> |x-6|.7-4=31

<=> |x-6|.7=35

<=> |x-6|=5

\(\Leftrightarrow\orbr{\begin{cases}x-6=5\\x-6=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=11\\x=1\end{cases}}}\)

please help me, please!!!!!!!!!!

a: \(x\in\left\{1;2;3;6;9;18\right\}\)

mà \(x\in B\left(4\right)\)

nên \(x\in\varnothing\)

b: \(x\in\left\{1;2;3;4;6;9;12;18;36\right\}\)

mà x>=12

nên \(x\in\left\{12;18;36\right\}\)

c: \(x\in\left\{0;12;24;36;48;60;72;84;96;108;...\right\}\)

mà 30<=x<=100

nên \(x\in\left\{36;48;60;72;84;96\right\}\)

d: \(x\inƯC\left(28;21\right)\)

\(\Leftrightarrow x\inƯ\left(7\right)\)

hay \(x\in\left\{1;7\right\}\)

ta có 2xy + x + y = 21

=> 4xy + 2x + 2y = 42

=> (4xy + 2x) + 2y = 42

=> 2x(2y+1) + 2y + 1 = 43

=> (2x + 1)(2y+1) = 43

\(\Rightarrow\hept{\begin{cases}2x+1\in\left\{1;43;-43;-1\right\}\\2y+1\in\left\{43;1;-1;-43\right\}\end{cases}}\Rightarrow\hept{\begin{cases}x\in\left\{0;21;-22;-1\right\}\\y\in\left\{21;0;-1;-22\right\}\end{cases}}\)

Vay bang bao nhieu the bn ? co ai b ko ?