\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

a, \(2.x^x=10.3^{12}+8.27^4\)

\(2.x^x=10.3^{12}+8.3^{12}\)

\(2.x^x=3^{12}.\left(10+8\right)\)

\(2.x^x=3^{12}.18\)

\(2.x^x=3^{12}.2.3^3\)

\(2.x^x=3^{15}.2\)

\(x^x=3^{15}\)( Hình như sai đề )

b,\(3^{2x+2}=9^{x+3}\)

\(3^{2x+2}=3^{2x+3}\)

câu b Sai đề

Số số hạng là :

      (2x - 2) : 2 + 1 = x - 1 + 1 = x (số)

Tổng là : 

       (2x + 2).x : 2 = 210

=> (2x² + 2x) : 2 = 210

=> x² + x = 210

=> x(x + 1) = 210

=> x(x + 1) = 20.21

=> x = 20

Vậy x = 20 

Ta có : \(\frac{x}{2}=\frac{10}{x+1}\)

=> x(x + 1) = 10.2

=> x(x + 1) = 20

=> sai đề 

a) (x + 1/2) . (2/3 − 2x) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

0,7.2\(\frac23\).20.0,375.\(\frac{5}{28}\)

= \(\frac{7}{10}\) .\(\frac83\).20.\(\frac38\).\(\frac{5}{28}\)

= (\(\frac{7}{10}\).\(\frac{5}{28}\).20).(\(\frac83\).\(\frac38\))

= (\(\frac18.20\)). 1

= \(\frac52\)

B = (9,75 .21\(\frac37\) + \(\frac{39}{4}\).18\(\frac47\))

B = (\(\frac{39}{4}\).\(\frac{150}{7}\frac{}{7}\)+ \(\frac{39}{4}\).\(\frac{130}{7}\))

B = \(\frac{39}{4}\).(\(\frac{150}{7}\) + \(\frac{130}{7}\))

B = \(\frac{39}{4}\).\(\frac{280}{7}\)

B = 390

a)  \(4^{2x-6}=1\)

\(\Rightarrow4^{2x-6}=4^0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

b) \(2^{x-1}=16\)

\(\Rightarrow2^{x-1}=2^4\)

\(\Rightarrow x-1=4\)

\(\Rightarrow x=5\)

c)  \(5< 5x< 125\)

\(\Rightarrow\frac{5}{5}< \frac{5x}{5}< \frac{125}{5}\)

\(\Rightarrow1< x< 25\)

\(\Rightarrow\left\{x\inℤ|1< x< 25\right\}\)

d) mk không hiểu

Câu 1:

Câu 1: x.(x+ 7) = 0

x = 0 hoặc x + 7 = 0

x + 7 = 0 suy ra x = - 7

Vậy x ∈ {0; - 7}

Câu 2: (x + 12).(x - 3) = 0

x + 12 = 0 hoặc x - 3 = 0

x + 12 = 0 suy ra x = - 12

x - 3 = 0 suy ra x = 3

Vậy x ∈ {-12; 3}

Câu 3:

(-x + 5).(3 - x) = 0

- x + 5 = 0; hoặc 3 - x = 0

- x + 5 = 0 suy ra x = 5

3 - x = 0 suy ra x = 3

Vậy x ∈ {0; 3}

Câu 4:

x(x + 2).(7 - x) = 0

x = 0 hoặc x + 2 = 0 hoặc 7 - x = 0

x + 2 = 0 suy ra x = - 2

7 - x = 0 suy ra x = 7

Vậy x ∈ {0; -2; 7}