\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)

\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)

\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)

\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)

Đkxđ: \(\hept{\begin{cases}x\ge-\frac{1}{4}\\y\ge2\end{cases}}\)

\(\Leftrightarrow2+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=y\Leftrightarrow2+\frac{1}{2}+\sqrt{x+\frac{1}{2}}=y\Leftrightarrow\sqrt{x+\frac{1}{2}}+\frac{5}{2}=y\)

do x,y nguyên dương nên \(\sqrt{x+\frac{1}{2}}+\frac{5}{2}\)nguyên dương\(\Leftrightarrow\sqrt{x+\frac{1}{2}}=\frac{k}{2}\)(K là số nguyên lẻ, \(k>1\))

\(\Rightarrow x=\frac{k^2-2}{4}\)

do \(k^2\)là số chính phương chia 4 dư 0,1 \(\Rightarrow x=\frac{k^2-2}{4}\notin Z\)

=> ko tồn tại cặp số nguyên dương x,y tmđkđb

\(\frac{42}{105}=\frac{2}{5}\)

=> \(\frac{x}{5}=\frac{4}{y}=\frac{z}{-80}=\frac{2}{5}\)

=> x=\(\frac{2}{5}\cdot5=2\)

y\(=4:\frac{2}{5}=4\cdot\frac{5}{2}=10\)

z=\(\frac{2}{5}\cdot\left(-80\right)=-32\)

tick nha

\(\Leftrightarrow\dfrac{16}{x+4}+\dfrac{16}{x-4}=\dfrac{5}{3}\)

=>\(\dfrac{16x-64+16x+64}{x^2-16}=\dfrac{5}{3}\)

=>5(x^2-16)=3*32x=96x

=>5x^2-96x-80=0

=>x=20 hoặc x=-4/5

Ta có : x/3=y/2      = x/12 = y /8 

         y/4=z/5       = y/8 = z/10 ( mình biến đổi sao cho y có mẫu chung là 8 ý bạn )

   => x/12=y/8=z/10 = -x-y+z/ -12-8+10 

                               = -10/-10 =1

=> x = 1.12=12

     y=1.8=8

    z=1.10=10

a)x/5=32/80

=>80x=32*5

x=160/8

x=20

b)13/x=26/30

=>26x=13*30

x=390/26

x=15

c)-x/7=22/-77

=>(-x)*(-77)=22*7

x*77=154

x=154/77

x=2

\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\left(1\right)\)

ĐKXĐ: \(x\ne\pm4\)

\(\left(1\right)\Leftrightarrow\frac{80\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{80\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)

\(\Leftrightarrow\frac{80x-320+80x+320}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)

\(\Leftrightarrow\frac{160x}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)

\(\Rightarrow480x=25\left(x-4\right)\left(x+4\right)\)

\(\Leftrightarrow480x=25\left(x^2-16\right)\)

\(\Leftrightarrow480x=25x^2-400\)

\(\Leftrightarrow25x^2-480x-400=0\)

\(\Leftrightarrow25x^2-480x+2304-2704=0\)

\(\Leftrightarrow\left(5x-48\right)^2=2704\)

\(\Leftrightarrow\left|5x-48\right|=52\)

  +Trường hợp 1: Nếu \(x\ge\frac{48}{5}\)thì \(5x-48\ge0\Rightarrow\left|5x-48\right|=5x-48\)

Ta có phương trình;

\(5x-48=52\)

\(\Leftrightarrow5x=100\)

\(\Leftrightarrow x=20\)(thỏa măn)

  +Trường hợp 2: Nếu\(x< \frac{48}{5}\)thì \(5x-48< 0\Rightarrow\left|5x-48\right|=48-5x\)

Ta có phương trình:

\(48-5x=52\)

\(\Leftrightarrow-5x=4\)

\(\Leftrightarrow x=-\frac{4}{5}\)(thỏa mãn)

\(S=\left\{20;-\frac{4}{5}\right\}\)

Ta có: \(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\)

    \(\Leftrightarrow80.\left[\frac{x-4+x+4}{\left(x+4\right).\left(x-4\right)}\right]=\frac{25}{3}\)

    \(\Leftrightarrow\frac{2x}{x^2-16}=\frac{25}{3}:80\)

    \(\Leftrightarrow\frac{2x}{x^2-16}=\frac{5}{48}\)

     \(\Rightarrow48.2x=5.\left(x^2-16\right)\)

    \(\Leftrightarrow96x=5x^2-80\)

    \(\Leftrightarrow5x^2-96x-80=0\)

    \(\Leftrightarrow\left(5x^2-100x\right)+\left(4x-80\right)=0\)

    \(\Leftrightarrow5x.\left(x-20\right)+4.\left(x-20\right)=0\)

    \(\Leftrightarrow\left(5x+4\right).\left(x-20\right)=0\)

    \(\Leftrightarrow\orbr{\begin{cases}5x+4=0\\x-20=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=-4\\x=20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\left(TM\right)\\x=20\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{-\frac{4}{5};20\right\}\)

Cách khác , mk cx chwua xm cách trên nhưng mk nghĩ là khác.

ĐKXĐ : \(\hept{\begin{cases}x+4\ne0\\x-4\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-4\\x\ne4\end{cases}}}\)

\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\)

\(\frac{80\left(x-4\right)3}{\left(x+4\right)\left(x-4\right)3}+\frac{80\left(x+4\right)3}{\left(x-4\right)\left(x+4\right)3}=\frac{25\left(x+4\right)\left(x-4\right)}{3\left(x+4\right)\left(x-4\right)}\)

\(80\left(x-4\right)3+80\left(x+4\right)3=25\left(x+4\right)\left(x-4\right)\)

\(240\left(x-4\right)+240\left(x+4\right)=25\left(x+4\right)\left(x-4\right)\)

\(480x=25x^2-400\)

\(480x-25x^2+400=0\)

\(5\left(96x-5x^2+80\right)=0\)

\(\left(5x+4\right)\left(x-20\right)=0\)

\(\orbr{\begin{cases}5x+4=0\\x-20=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-4\\x=20\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{4}{5}\\x=20\end{cases}}}\)

Theo ĐKXĐ : => x (TM)