Đặt \(\frac{x-18}{x+4}=\frac{x-17}{x+16}=k\)

Suy ra: \(x-18=k\left(x+4\right)\Rightarrow x=\frac{4k+18}{1-k}\left(1\right)\\ x-17=k\left(x+16\right)\Rightarrow x=\frac{16k+17}{1-k}\left(2\right)\)

Từ (1) và (2) ta được: \(4k+18=16k+17,\) suy ra \(k=\frac{1}{12},x=20\)

\(\frac{x+15}{2000}+\frac{x+16}{1999}=\frac{x+17}{1998}+\frac{x+18}{1997}\)

\(\Leftrightarrow\frac{x+15}{2000}+1+\frac{x+16}{1999}+1=\frac{x+17}{1998}+1+\frac{x+18}{1997}+1\)

\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}=\frac{x+2015}{1998}+\frac{x+2015}{1997}\)

\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}-\frac{x+2015}{1998}-\frac{x+2015}{1997}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\right)=0\)

Có: \(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\ne0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

nhớ là có ai làm rồi mà quên =))

khá dài à =)) bấm máy mỏi :v

\(a,\frac{x}{3}=\frac{y}{5}\)

=> (x+y)/(3+5) = x/3 = y/5

=> 32/8 = x/3 = y/5

=> 4 = x/3 = y/5

=> x = 12 ; y = 20

b, x/2 = y/5 

=> x + y/2 + 5 = x/2 = y/5

=> 21/7 = x/2 = y/5

=> 3 = x/2 = y/5

=> x = 6 và y = 15

c.7x=3y và x-y=16

đặt x/3=y/7

=>x/3=y/7=x-y/3-7=16/-4=-4(vì x-y=16)

=>x/3=-4=-12

=>y/7=-4=-28

vậy .....

Ta có: \(\frac{x+y}{16}=\frac{x-y}{18}\)

=> 18(x + y) = 16(x - y)

=> 18x + 18y = 16x - 16y

=> 18x - 16x = -16y - 18y

=> 2x = -34y

=> x = -17y

Khi đó: \(\frac{-17y+y}{16}=\frac{-17y.y}{17}\)

=> \(\frac{-16y}{16}=-y^2\)

=> \(-y+y^2=0\)

=> y(y - 1) = 0

=> \(\orbr{\begin{cases}y=0\\y-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

Với y = 0 => x = -17.0 = 0

   y=  1 => x = -17 . 1 = -17

Vậy ....

\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)

\(\Rightarrow\frac{x+18}{2018}-1+\frac{x+17}{2017}-1+\frac{x+16}{2016}-1=3-3\)

\(\Rightarrow\frac{x+18-2018}{2018}+\frac{x+17-2017}{2017}+\frac{x+16-2016}{2016}=0\)

\(\Rightarrow\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)

=> x - 2000 = 0

=> x            = 2000

Ta có : 

\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)

\(\Leftrightarrow\)\(\left(\frac{x+18}{2018}-1\right)+\left(\frac{x+17}{2017}-1\right)+\left(\frac{x+16}{2016}-1\right)=3-3\) ( trừ hai vế cho 3 ) 

\(\Leftrightarrow\)\(\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)

\(\Leftrightarrow\)\(\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)

Nên \(x-2000=0\)

\(\Rightarrow\)\(x=2000\)

Vậy \(x=2000\)

Chúc bạn học tốt ~ 

Xét \(\frac{-5}{x}=\frac{-18}{72}\)

\(\Rightarrow\)(-5) . 72 = x . (-18)

\(\Rightarrow\)-360 = x . (-18)

\(\Rightarrow\)x = (-360) : (-18) = 20

Xét \(\frac{y}{16}=\frac{-18}{72}\)

\(\Rightarrow\)72y = 16 . (-18)

\(\Rightarrow\)72y = -288

\(\Rightarrow\)y = (-288) : 72 = -4

Vậy x = 20 ; y = -4

Ta có : 

   \(\frac{-5}{x}=\frac{-18}{72}\Rightarrow\frac{-5}{x}=\frac{-1}{4}\)  

                            \(\Rightarrow-5.4=x.-1\) 

                            \(\Rightarrow-20=-x\)

                            \(\Rightarrow x=20\)

Thay x = 20 vào  \(\frac{-5}{x}\)ta được : 

\(\frac{-5}{20}=\frac{y}{16}\)\(\Rightarrow\frac{-1}{4}=\frac{y}{16}\Rightarrow-1.16=y.4\)

                                                   \(\Rightarrow-16=y.4\)  

                                                    \(\Rightarrow y=-4\)  

Vậy x = 20 , y = -4

Tk mk nha !!! 

Ta có: 1/2 - (1/3 + 1/4) = 1/2 - 7/12 = -1/12 ;

           1/48 - (1/16 - 1/6) = 1/48 + 5/48 = 1/8

Vì \(-\frac{1}{12}< x< \frac{1}{8}\) nên x = 0

Ta có :

\(4\frac{5}{9}:2\frac{5}{18}-7=2-7=-5\)

\(\left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)=\left(1+\frac{38}{5}\right):\left(-21\frac{2}{3}\right)=\frac{43}{5}:\frac{-65}{3}=-\frac{129}{325}\)

Vì \(-5< x< -\frac{129}{325}\) nên \(x\in\left\{-4;-3;-2;-1\right\}\)

no l can't help you

1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)

\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu

\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)

\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)

Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)

Bài 1b) có thể giải gọn hơn nhuư thế này

Cách gọn:

1b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\)trái dấu

Mà \(x^2+4\ge0\) nên \(\left|x\right|-3< 0\Leftrightarrow\left|x\right|< 3\)

\(\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)