tui o bít nhưng ai kb vs tui o

a)x=30

b)x=65

a) x-\(\left(\frac{50x}{100}+\frac{25x}{200}\right)\)=\(11\frac{1}{4}\)

<=>x - \(\frac{1}{2}x-\frac{1}{8}x\)=\(\frac{11.4+1}{4}\)

<=>\(\frac{3}{8}x=\frac{45}{4}\)

<=>x=\(\frac{45}{4}:\frac{3}{8}\)

<=>x=30

Vậy x=30

c) (x-5).\(\frac{30}{100}\)=\(\frac{20x}{100}+5\)

<=>(x-5).\(\frac{3}{10}\)=\(\frac{x}{5}\)+5

<=>\(\frac{3x}{10}-\frac{15}{10}=\frac{x}{5}+5\)

<=>\(\frac{3x}{10}-\frac{x}{5}=5+\frac{15}{10}\)

<=>\(\frac{x}{10}=\frac{13}{2}\)

<=>x=\(\frac{13.10}{2}\)

<=>x=65

Vậy: x=65

 \(x-\left(\frac{50x}{100}+\frac{25x}{100}\right)=\frac{45}{4}\)

      \(x-\left(\frac{50x+25x}{100}\right)=\frac{45}{4}\)

                    \(x-\frac{75x}{100}=\frac{45}{4}\)

               \(x-x\times\frac{3}{4}=\frac{45}{4}\)

         \(x\times\left(1-\frac{3}{4}\right)=\frac{45}{4}\)

                       \(x\times\frac{1}{4}=\frac{45}{4}\)

                                 \(x=\frac{45}{4}\div\frac{1}{4}\)

                                 \(x=45\)

\(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=\frac{45}{4}\)

\(x-\left(\frac{100x}{200}+\frac{25x}{200}\right)=\frac{45}{4}\)

\(x-\left(\frac{100x+25x}{200}\right)=\frac{45}{4}\)

\(x-\frac{125x}{200}=\frac{45}{4}\)

\(x-x\frac{5}{8}=\frac{45}{4}\)

\(x\left(1-\frac{5}{8}\right)=\frac{45}{4}\)

\(x.\frac{3}{8}=\frac{45}{4}\)

\(x\)=\(\frac{45}{4}:\frac{3}{8}\)

\(x\)=\(30\)

uk

\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\Rightarrow x=305\)

\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)

\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)

\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)

\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)

\(\Rightarrow3x=\frac{45}{4}\cdot8\)

\(\Rightarrow3x=90\Rightarrow x=30\)

\(c)1+2+3+4+...+x=820\)

Ta có : \(1+2+3+4+...+x=\frac{(1+x)\cdot x}{2}\)

Do đó : \(\frac{(1+x)\cdot x}{2}=820\)

\(\Rightarrow(1+x)\cdot x=820\cdot2\)

\(\Rightarrow(1+x)\cdot x=1640\)

\(\Rightarrow(1+x)\cdot x=40\cdot41\)

Vì x và x + 1 là hai số tự nhiên liên tiếp nên => x = 40

Chúc bạn học tốt :3

1+1=3 :)))

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).