\(\frac{x-18}{x+4}\)= \(\frac{x-17}{x+16}\)
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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 \(k=\frac{1}{12},x=20\)
Hôm nay cô giao bài nhìu, tui đăng nhìu, vất vả nhìu cho chú đấy
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
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
\(\frac{x+15}{2000}+\frac{x+16}{1999}=\frac{x+17}{1998}+\frac{x+18}{1997}\)
\(\Leftrightarrow\frac{x+15}{2000}+1+\frac{x+16}{1999}+1=\frac{x+17}{1998}+1+\frac{x+18}{1997}+1\)
\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}=\frac{x+2015}{1998}+\frac{x+2015}{1997}\)
\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}-\frac{x+2015}{1998}-\frac{x+2015}{1997}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\right)=0\)
Có: \(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\ne0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
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 ~
\(\frac{x-18}{x+4}=\frac{x-17}{x+16}\)
\(\Rightarrow\left(x-18\right).\left(x+16\right)=\left(x+4\right).\left(x-17\right)\)
\(x^2+16x-18x-288=x^2-17x+4x-68\)
\(x^2-2x-288=x^2-13x-68\)
\(\Rightarrow x^2-2x-x^2+13x=-68+288\)
\(11x=220\)
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
\(\frac{x-18}{x+4}=\frac{x-17}{x+16}\Leftrightarrow\left(x-18\right)\left(x+16\right)=\left(x-17\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+16x-18x-18\cdot16=x^2+4x-17x-17\cdot4\)
\(\Leftrightarrow-2x-288=-13x-68\Leftrightarrow-2x+13x=-68+288\)
\(\Leftrightarrow11x=220\Leftrightarrow x=\frac{220}{11}=20\)