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a) \(x^3\)+\(x^2\)=36
\(\Leftrightarrow\)\(x^3\)+\(x^2\)\(-36=0\)
\(\Leftrightarrow\)\(x^3\)\(-3x^2\)\(+4x^2\)\(-12x\)\(+12x-36=0\)
\(\Leftrightarrow\)\(x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2+4x+12\right)=0\)
Suy ra: \(x-3=0\) hoặc \(x^2+4x+12=0\)
- \(x-3=0\) \(\Leftrightarrow\) \(x=3\)
- \(x^2+4x+12=0\) (phương trình vô nghiệm)
Vậy \(x=3\)
\(=\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=\left(x+1\right)\left(x^2-7x+19\right)=0\)
Ta thấy: \(x^2-7x+19=x^2-2\times\frac{7}{2}x+\frac{7}{2}^2+\frac{27}{4}=\left(x-\frac{7}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)lớn hơn 0
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(x^3-6x^2+12x+19=0\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-7x+19\right)=0\)
Mà \(x^2-7x+19>0\)với \(\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1\)
\(\dfrac{1}{2}x^3-49x=0\)
\(\dfrac{1}{2}x^2.x-49x=0\)
\(x.\left(\dfrac{1}{2}x^2-49\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2-49=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2=49\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=98\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=7\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0,7\sqrt{2}\right\}\)
49. x3-x=0
<=> 49. x(x2-1)=0
<=> 49. x(x2-12)=0
<=> 49.x(x-1)(x+1)=0
=> x=0 hoặc x-1=0 hoặc x+1=0
=> x=0 hoặc x=1 hoặc x= -1
49. x3-x=0
<=> 49. x(x2-1)=0
<=> 49. x(x2-12)=0
<=> 49.x(x-1)(x+1)=0
=> x=0 hoặc x-1=0 hoặc x+1=0
=> x=0 hoặc x=1 hoặc x= -1
a, \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\Leftrightarrow\left(4x-3x-3\right)\left(4x+2x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)
c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)
1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)
\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3: Ta có: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
\(\hept{\begin{cases}x=7\\x=-7\\x=0\end{cases}}\)
Ta có : x3 - 49x = 0
=> x(x2 - 49) = 0
=> x(x - 7)(x + 7) = 0
<=> x = 0
x - 7 = 0
x + 7 = 0
<=> x = 0
x = 7
x = -7