(x-4)^2-36=0
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1, \(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
vậy \(x\) \(\in\) {-3; 3}
5, 4\(x^2\) - 36 = 0
4.(\(x^2\) - 9) = 0
\(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; 3}
`@` `\text {Ans}`
`\downarrow`
`1,`
`x^2 - 9 = 0`
`<=> x^2 = 0 + 9`
`<=> x^2 = 9`
`<=> x^2 = (+-3)^2`
`<=> x = +-3`
Vậy, `S = {3; -3}`
`2,`
`25 - x^2 = 0`
`<=> x^2 = 25 - 0`
`<=> x^2 = 25`
`<=> x^2 = (+-5)^2`
`<=> x = +-5`
Vậy,` S= {5; -5}`
`3,`
`-x^2 + 36 = 0`
`<=> -x^2 = 0 - 36`
`<=> -x^2 = -36`
`<=> x^2 = 36`
`<=> x^2 = (+-6)^2`
`<=> x = +-6`
Vậy, `S= {6; -6}`
`4,`
`4x^2 - 4 = 0`
`<=> 4x^2 = 0+4`
`<=> 4x^2 = 4`
`<=> x^2 = 4 \div 4`
`<=> x^2 = 1`
`<=> x^2 = (+-1)^2`
`<=> x = +-1`
Vậy, `S= {1; -1}`
`@` `\text {Kaizuu lv uuu}`
a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252
a, x\(^2\) - x = x - 1
\(\Leftrightarrow\) x\(^2\) - 2x + 1 = 0
\(\Leftrightarrow\) (x - 1)\(^2\) = 0
\(\Leftrightarrow\) x - 1 = 0
\(\Leftrightarrow\) x = 1
a) \(x^2-x=x-1\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Rightarrow x=1\)
b) \(\left(x^2-36\right)-\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
Vậy..
c) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow-4x+2=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
d) \(x^2\left(x^2-4\right)-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm1\end{matrix}\right.\)
Vậy..
\(4\left(6-x\right)+x^2-12x+36=0\)
\(24-4x+x^2-12x+36=0\)
\(x^2-16x+60=0\)
\(x^2-2x8+8^2-8^2+60=0\)
\(\left(x-8\right)^2-4=0\)
\(\left(x-8\right)^2=4\)
\(\left(x-8\right)^2=\left(\pm2\right)^2\)
\(\orbr{\begin{cases}x-8=2\Rightarrow x=10\\x-8=-2\Rightarrow x=6\end{cases}}\)
a) \(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)
b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)
\(x^3-4x^2-9x+36=0\)
\(x^2\left(x-4\right)-9\left(x-4\right)=0\)
\(\left(x-4\right)\left(x^2-9\right)=0\)\(\)
\(\Rightarrow x-4=0\) hay \(x^2-9=0\)
\(\Rightarrow x=4\) hay \(x^2=9=3^2\)
\(\Rightarrow x=4\) hay \(x=\pm3\)
⇔x2(x-4) -9(x-4) = 0
⇔(x-4).(x-3).(x+3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
`x-4` mũ `2=36`
TH1
`x-4=6`
`x=10`
TH2
`x-4=-6`
`x=-2`
Vậy `x=10` hoặc `x=-2`
(x - 4)2 - 36 = 0
(x - 4)2 = 0 + 36
(x - 4)2 = 36
(x - 4)2 = 62 = (-6)2
TH1 : (x - 4)2 = 62
x - 4 = 6
x = 6 + 4
x = 10
TH2 : (x - 4)2 = (-6)2
x - 4 = -6
x = -6 + 4
x = -2
vậy x ϵ {-2;10}