\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

a) \(\frac{x+7}{x+4}=\frac{2}{5}\)

\(\Rightarrow5\left(x+7\right)=2\left(x+4\right)\)

\(\Rightarrow5x+35-2x-8=0\)

\(\Rightarrow3x=-27\)

\(\Rightarrow x=-9\)

b) \(\frac{2x-3}{2}=\frac{50}{2x-3}\)

\(\Rightarrow\left(2x-3\right)^2=100\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-3=10\\2x-3=-10\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{13}{2}\\x=-\frac{7}{2}\end{array}\right.\)

c) \(\frac{x+1}{x-3}=\frac{x+3}{x+2}\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)=\left(x-3\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+3x+2=x^2-9\)

\(\Leftrightarrow3x=-11\)

\(\Leftrightarrow x=-\frac{11}{3}\)

c) Tìm các số nguyên x,y thỏa mãn

*\(2xy+6x-y=10\)

\(\Leftrightarrow\left(2xy+6x\right)-y-3=10-3=7\)

\(\Leftrightarrow2x\left(y+3\right)-\left(y+3\right)=7\)

\(\Leftrightarrow\left(y+3\right)\left(2x-1\right)=7\)

Lập bảng xét ước nữa là xong.

* \(xy+4x-3y=1\Leftrightarrow\left(xy+4x\right)-3y-12=1-12=-11\)

\(\Leftrightarrow x\left(y+4\right)-\left(3y+12\right)=-11\)

\(\Leftrightarrow x\left(y+4\right)-3\left(y+4\right)=-11\)

\(\Leftrightarrow\left(x-3\right)\left(y+4\right)=-11\)

Lập bảng xét ước nữa là xong.

Mới nhìn vào thấy bài toán hay hay lạ kì.

Thêm một vào bớt một ra

Tức thì bài toán trở nên dễ dàng:

 \(\frac{x}{50}-\frac{x-1}{51}=\frac{x+2}{48}-\frac{x-3}{53}\) 

\(\Leftrightarrow\frac{x}{50}+1-\frac{x-1}{51}-1=\frac{x+2}{48}+1-\frac{x-3}{53}-1\)

\(\Leftrightarrow\left(\frac{x}{50}+1\right)-\left(\frac{x-1}{51}+1\right)=\left(\frac{x+2}{48}+1\right)-\left(\frac{x-3}{53}+1\right)\)

\(\Leftrightarrow\frac{x+50}{50}-\frac{x+50}{51}=\frac{x+50}{48}-\frac{x+50}{53}\)

\(\Leftrightarrow\frac{x+50}{50}-\frac{x+50}{51}-\frac{x+50}{48}+\frac{x+50}{53}=0\)

\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{50}-\frac{1}{51}-\frac{1}{48}+\frac{1}{53}\right)=0\)

Dễ thấy \(\left(\frac{1}{50}-\frac{1}{51}-\frac{1}{48}+\frac{1}{53}\right)\ne0\)

Do đó x + 50 = 0 hay x = -50

a,\(\frac{x+1}{2019}+1+\frac{x}{1010}+2+\frac{x-2}{674}+3+\frac{x-4}{506}+4=0\)

\(\frac{x+2020}{2019}+\frac{x+2020}{1010}+\frac{x+2020}{674}+\frac{x+2020}{506}=0\)

\(\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{1010}+\frac{1}{674}+\frac{1}{506}\right)=0\)

Vì \(\left(\frac{1}{2019}+\frac{1}{1010}+\frac{1}{674}+\frac{1}{506}\right)\ne0\)

\(x+2020=0\Rightarrow x=-2020\)

Vậy...

a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)

<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{\left(x-1\right)-\left(y-2\right)+\left(z-3\right)}{2-3+4}\)\(=\frac{x-1-y+2+z-3}{3}=\frac{50-2}{3}=\frac{48}{3}=16\)

\(\Rightarrow\hept{\begin{cases}x=16.2+1=33\\y=16.3+2=50\\z=16.4+3=67\end{cases}}\)

Vậy ........................

b) \(\frac{x-1}{2}=\frac{2x-2}{4}\)

\(\frac{y-2}{3}=\frac{3y-6}{9}\)

\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z+3-2-6}{9}=\frac{50+3-2-6}{9}=\frac{45}{9}=5\)=>x-1=5.2=10

=>x=11

y-2=5.3=15

=>y=17

z-3=5.4=20

=>z=23

Vậy (x;y;z)=(11;17;23)

Áp dụng t/c của dãy tỉ số bằng nhau:

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+x-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(vì x+y+z khác 0).Do đó x+y+z = 0.5

Thay kq này vào bài ta được:

\(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)

Tức là : \(\frac{1,5-x}{x}=\frac{2,5-y}{y}=\frac{-2,5-z}{z}=2\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\)