Tìm x
(8.x - 16).(x -5)=0
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Vì \(\left(8\times x-16\right)\times\left(x-5\right)=0\)
=> \(8\times x-16=0\) hoặc \(x-5=0\)
TH1:
\(8\times x-16=0\)
\(8\times x=16\)
\(x=16\div8\)
\(x=2\)
TH2:
\(x-5=0\)
\(x=5\)
Vậy \(x\in\left\{2;5\right\}\)
(8 - 16)(\(x\) - 5) = 0
\(x\) - 5 = 0
\(x\) = 5
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)
\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)
\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)
Vậy x = 5/19
\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy x = 1/2 hoặc x = -6
\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Vậy x = 7 hoặc x = -1
`1-(5 3/8+x-7 5/24):(-16 2/3)=0`
`=>1-(5+3/8+x-7-5/24):(-50/3)=0`
`=>1=(x-2-11/6):(-50/3)`
`=>1=(x-11/6):(-50/3)`
`=>x-11/6=-50/3`
`=>x=-89/6`
Vậy `x=-89/6`
\(1-\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow1-\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right):\dfrac{50}{3}=0\)
\(\Leftrightarrow1-\left(\dfrac{129}{24}-\dfrac{173}{24}+x\right).\dfrac{3}{50}=0\)
\(\Leftrightarrow1+\dfrac{11}{6}-x.\dfrac{3}{50}=0\)
\(\Leftrightarrow\dfrac{17}{6}-x.\dfrac{3}{50}=0\)
\(\Leftrightarrow x.\dfrac{3}{50}=\dfrac{17}{6}\)
\(\Leftrightarrow x=\dfrac{425}{9}\)
-Chúc bạn học tốt-
\(\text{ a, 5-(10x)=-7}\)
\(\Rightarrow\) 10x=5-(-7)
\(\Rightarrow\) 10x=12
\(\Rightarrow\) x=12:10
\(\Rightarrow\) x=1,2
b, (x-5).(2x+8)=0
\(\Rightarrow\) x-5=0 hoặc 2x+8=0
\(\Rightarrow\) x =0+5 \(\Rightarrow\) 2x =0+8
\(\Rightarrow\) x =5 \(\Rightarrow\) 2x =8
\(\Rightarrow\) x =8:2
\(\Rightarrow\) x =4
vậy x\(\in\){5;4}
c, 2x-9=-8+9
\(\Rightarrow\) 2x-9=1
\(\Rightarrow\) 2x =1+9
\(\Rightarrow\) 2x =10
\(\Rightarrow\) x =10:2
\(\Rightarrow\) x =5
d, |x-9|.(-8)=-16
\(\Rightarrow\)|x-9| =-16:(-8)
\(\Rightarrow\)|x-9| =2
\(\Rightarrow\) x-9 =\(\hept{\begin{cases}2\\-2\end{cases}}\)
trường hợp 1: x-9=2
\(\Rightarrow\) x =2+9
\(\Rightarrow\) x =11
trường hợp 2: x-9=-2
\(\Rightarrow\) x =-2+9
\(\Rightarrow\) x =7
vậy x \(\in\){11;7}
# học tốt #
`(8x-16).(x-5)=0`
`=>8x-16=0` hoặc `x-5=0`
`=>8x=16` hoặc `x=5`
`=>x=2` hoặc `x=5`
x = 5