\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\Leftrightarrow\left(\dfrac{x-45}{55}-1\right)+\left(\dfrac{x-47}{53}-1\right)=\left(\dfrac{x-55}{45}-1\right)+\left(\dfrac{x-53}{47}-1\right)\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)

Do \(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\) nên x - 100 = 0 <=> x = 100

\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)

\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)

\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)

\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)

\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)

Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)

\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)

\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)

\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)

\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)

Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)

\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)

\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)

\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)

\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)

\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)

Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)

Ta có : \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{49}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)

\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{49}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)

\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{49}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)

<=> x - 100 = 0

<=> x = 100

Vậy ..

Ta có: \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)

\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{48}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)

\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{48}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)

mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\)

nên x-100=0

hay x=100

Vậy: S={100}

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

=>\(\dfrac{3}{2}\sqrt{x-3}=3\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

Thêm 2 vào pt có :

\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)                (1)

\(\Leftrightarrow\frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1\)

\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\) (2)

\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\ne0\)

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

k mk nha!

thanks

thanks

\(\dfrac{180}{x-4}-\dfrac{180}{x}=\dfrac{1}{2}\)

\(\Leftrightarrow\) \(\dfrac{2x\cdot180}{2x\left(x-4\right)}-\dfrac{2\cdot180\cdot\left(x-4\right)}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{360x-360x+1440-x^2+4x}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{-x^2+4x+1440}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow-x^2+4x+1440=0\)

\(\Leftrightarrow-x^2+40x-36x+1440=0\)

\(\Leftrightarrow-x\cdot\left(x-40\right)\cdot\left(-36\right)\cdot\left(x-40\right)=0\)

\(\Leftrightarrow\left(x-40\right)\cdot\left(x-36\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-40=0\\x+36=0\end{matrix}\right.\)

 \(x-40=0\)

  \(x=0+40\)

 \(x=40\)

\(x+36=0\)

   \(x=0-36\)

   \(x=-36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)

\(180\left(\dfrac{1}{x-4}-\dfrac{1}{x}\right)=\dfrac{1}{2}\)

\(\dfrac{1}{x-4}-\dfrac{1}{x}=\dfrac{1}{360}\left(đk:x\ne0,4\right)\)

\(\dfrac{x-x+4}{x\left(x-4\right)}=\dfrac{1}{360}\)

\(\dfrac{4}{x\left(x-4\right)}=\dfrac{1}{360}\)

\(x^2-4x=1440\)

\(x^2-4x+4=1444\)

\(\left(x-2\right)^2=1444=38^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=38\\x-2=-38\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)