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 ....

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

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

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

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

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

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 =))

Dễ lắm bạn ạ 

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

\(\Leftrightarrow\left(x-18\right)\left(x+16\right)=\left(x+4\right)\left(x-17\right)\)

\(\Leftrightarrow x^2+16x-18x-288=x^2+4x-17x-68\)

\(\Leftrightarrow x^2-2x-288=x^2-13x-68\)

\(\Leftrightarrow x^2-x^2-2x+13x=-68+288\)

\(\Leftrightarrow11x=220\)

\(\Leftrightarrow x=20\)

(x-18)(x+16)=(x+4)(x-17)

x²-2x--288=x²-13x-68

x²-x²-2x+13x-288+68=0

11x=220

x=220:11

x=20

no l can't help you

a, \(\frac{x}{y+z+1}=\frac{y}{x+z+3}=\frac{z}{x+y-4}=\frac{x+y+z}{y+z+1+x+z+3+x+y-4}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

=>\(x+y+z=\frac{1}{2};\frac{x}{y+z+1}=\frac{1}{2};\frac{y}{x+z+3}=\frac{1}{2};\frac{z}{x+y-4}=\frac{1}{2}\)

=>\(\hept{\begin{cases}y+z+1=2x\\x+z+3=2y\\x+y-4=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+3=3y\\x+y+z-4=3z\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+3\\3z=\frac{1}{2}-4\end{cases}}}\Rightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{7}{2}\\3z=\frac{-7}{2}\end{cases}}\)

đến đây dễ rồi

b, =>(x-18)(x+16)=(x+4)(x-17)

=>x²+16x-18x-288=x²-17x+4x-68

=>x²-2x-288-x²+13x+68=0

=>11x-220=0

=>11x=220

=>x=20

\(\frac{x-18}{x+4}=\frac{x-17}{x+16}\Leftrightarrow\left(x-18\right)\left(x+16\right)=\left(x-17\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+16x-18x-18\cdot16=x^2+4x-17x-17\cdot4\)
\(\Leftrightarrow-2x-288=-13x-68\Leftrightarrow-2x+13x=-68+288\)
\(\Leftrightarrow11x=220\Leftrightarrow x=\frac{220}{11}=20\)

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

\(\Rightarrow\left(x-18\right).\left(x+16\right)=\left(x+4\right).\left(x-17\right)\)

\(x^2+16x-18x-288=x^2-17x+4x-68\)

\(x^2-2x-288=x^2-13x-68\)

\(\Rightarrow x^2-2x-x^2+13x=-68+288\)

\(11x=220\)

x = 220:11

x = 20

      x - 18 / x + 4 = x -17 / x + 16

   <=> (x-18) . ( x+ 16) = (x+4) . (x-17)

<=> x² + 16x - 18x -288 = x² -17x + 4x -68

<=>x² - 2x -288 = x² -13x -68

<=> x² - x² -2x + 13x = 288 - 68

<=> 11x = 220 => x =  20 

vậy x= 20 

chúc bạn hok tốt và nhớ ủng hộ mik nha

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