Tìm x
(8x-16):(x-5)=0
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`(8x-16)(x-5)=0`
`=>8x-16=0` hoặc `x-5=0`
`=>8x=16` hoặc `x=5`
`=>x=16:8` hoặc `x=5`
`=>x=2` hoặc `x=5`
Vậy `x in{2;5}`
TH1: 8x-16=0 8x =0+16 8x =16 x =16:8 x =2 | TH2: x-5=0 x=0+5 x=5 |
\(\left(8x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
8x -16(x-5)=0
8x-16x+80=0
8x - 16x=0-80
x.(8-16)=-80
x.(-8)=-80
x=-80:(-8)
x=10
Vậy x=10
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
(8x-16)(x-5)=0
=>8x-16=0 hoặc x-5=0
=>x=2 hoặc x=5.
Chúc bạn học tốt nhé
\(\left(8x-16\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`a, x^2-10x+25=0`
`<=>x^2 -2.x.5+5^2=0`
`<=>(x-5)^2=0`
`<=>x-5=0`
`<=>x=5`
__
`x^2 -8x+16=0`
`<=> x^2 - 2.x.4+4^2=0`
`<=>(x-4)^2=0`
`<=>x-4=0`
`<=>x=4`
__
`x^2-49=0`
`<=>x^2 - 7^2=0`
`<=>(x-7)(x+7)=0`
`<=>x-7=0` hoặc `x+7=0`
`<=> x=7` hoặc `x=-7`
__
`4x^2-25=0`
`<=> (2x)^2 -5^2=0`
`<=>(2x-5)(2x+5)=0`
`<=>2x-5=0` hoặc `2x+5=0`
`<=> 2x=5` hoặc `2x=-5`
`<=>x=5/2` hoặc `x=-5/2`
a: =>(x-5)^2=0
=>x-5=0
=>x=5
b: =>(x-4)^2=0
=>x-4=0
=>x=4
c: =>(x-7)(x+7)=0
=>x-7=0 hoặc x+7=0
=>x=7 hoặc x=-7
d: =>(2x-5)(2x+5)=0
=>2x-5=0 hoặc 2x+5=0
=>x=5/2 hoặc x=-5/2
\(\Rightarrow\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(\left(8x-16\right)\left(x-5\right)=0\\ \Leftrightarrow8\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
(8\(x\) - 16).(\(x\) - 5) = 0
\(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}8x=16\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy \(x\) \(\in\){2; 5}