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

(x-5) x \(\frac{3}{10}\)= 200x : 100 + 5

\(\frac{x.3-15}{10}\)= 2x + 5

\(\frac{x.3-15}{10}-2x=5\)

\(\frac{x.3-15}{10}-\frac{20x}{10}=5\)

\(\frac{3x-15-20x}{10}=5\)

3x - 20x - 15 = 50

-17x - 15 = 50

-17x = 65

x = 65 : -17

x = \(\frac{-65}{17}\)

a) \(2^{x+2}-2^2=96\)

 <=> \(2^x.2^2-2^x=96\)

 <=> \(2^x\left(4-1\right)=96\)

<=> \(3.2^x=96\)

<=> \(2^x=32\)

<=> \(2^x=2^5\)

<=> x = 5 

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

\(\Rightarrow x-\left(\frac{1x}{2}+\frac{1x}{8}\right)=\frac{45}{4}\)

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

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

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

\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\Rightarrow x=\frac{45}{4}\div\frac{3}{8}=30\)

Vậy x = 30 

a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)

=>x+2006=0

=>x=-2006

b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)

=>x-105=0

=>x=105

b: \(\Leftrightarrow\left(\dfrac{29-x}{21}+1\right)+\left(\dfrac{27-x}{23}+1\right)+\left(\dfrac{25-x}{25}+1\right)+\left(\dfrac{23-x}{27}+1\right)+\left(\dfrac{21-x}{29}+1\right)=0\)

=>50-x=0

hay x=50

c: \(\Leftrightarrow\dfrac{x-2}{2001}+1=\dfrac{x-1}{2002}+\dfrac{x}{2003}\)

\(\Leftrightarrow\left(\dfrac{x-2}{2001}-1\right)=\left(\dfrac{x-1}{2002}-1\right)+\left(\dfrac{x}{2003}-1\right)\)

=>x-2003=0

hay x=2003

\(< =>\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)+\left(\dfrac{x-3}{102}-1\right)+3=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)+\left(\dfrac{x-102}{3}-1\right)+3\)\(< =>\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}\)

\(< =>\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\) = 0

<=> x - 105 = 0

<=> x = 105

Vậy tập nghiệm của phương trình là S = \(\left\{105\right\}\)

=) đây là bài giải bằng cách lập pt mà nãy bạn đã đăng nè:v mà giải thì ra vô nghiệm á bạn nên mik ko có làm:v

\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{30}{60}=0,5\left(ĐKXĐ:x\ne0;x\ne-10\right)\\ \Leftrightarrow\dfrac{100\left(x+10\right)-100x}{x\left(x+10\right)}=\dfrac{0,5x\left(x+10\right)}{x\left(x+10\right)}\\ \Leftrightarrow100x-100x+1000=0,5x^2+5x\\ \Leftrightarrow0,5x^2+5x-1000=0\\ \Leftrightarrow0,5x^2-20x+25x-1000=0\\ \Leftrightarrow0,5x.\left(x-40\right)+25.\left(x-40\right)=0\\ \Leftrightarrow\left(0,5x+25\right)\left(x-40\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,5x+25=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-50\\x=40\end{matrix}\right.\\ Vậy:S=\left\{-50;40\right\}\)