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

tui o bít nhưng ai kb vs tui o

\(c,\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow(x+1)(x+1)=2.8\)

\(\Rightarrow(x+1)^2=16\)

\(\Rightarrow(x+1)^2=4^2\)

\(\Rightarrow x+1=4\)

\(\Rightarrow x=4-1\)

\(\Rightarrow x=3\)

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

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

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

\(\Rightarrow\frac{100x-75x}{100}=\frac{1125}{100}\)

\(\Rightarrow25x=1125\)

\(\Rightarrow x=45\)

\(b,(x-5)\frac{30}{100}=\frac{200x}{100}+5\)

\(\Rightarrow\frac{30x}{100}-\frac{3}{2}=\frac{200x}{100}+5\)

\(\Rightarrow\frac{30x}{100}-\frac{200x}{100}=5+\frac{3}{2}\)

\(\Rightarrow\frac{-170x}{100}=\frac{13}{2}\)

\(\Rightarrow\frac{-170x}{100}=\frac{650}{100}\)

\(\Rightarrow-170x=650\)

\(\Rightarrow x=\frac{-65}{17}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

uk

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

\(x-\dfrac{3x}{4}=\dfrac{45}{4}\)

\(\dfrac{x}{4}=\dfrac{45}{4}\Rightarrow x=45\)

Ta có: \(x-\left(\dfrac{50x}{100}+\dfrac{25x}{200}\right)=11\dfrac{1}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}x-\dfrac{1}{8}x=\dfrac{45}{4}\)

\(\Leftrightarrow x\cdot\dfrac{3}{8}=\dfrac{45}{4}\)

hay x=30

đỡ hơn chưa??? mong các bn giúp mình vs

Vê trái: 

\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)

\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)

\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)

Vế phải:

\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)

\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải

=> đpcm