a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
*Cách khác:
a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)
Bài 5:
b) Ta có: \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow x^3-\dfrac{1}{49}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{49}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{7}\right)\left(x+\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{7}=0\\x+\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\\x=-\dfrac{1}{7}\end{matrix}\right.\)
Vậy: \(S=\left\{0;\dfrac{1}{7};-\dfrac{1}{7}\right\}\)
a) (3x-1)2-(x-3)2=0
ĐKXĐ: \(|^{3x-1\ne0}_{x-3\ne0}\Leftrightarrow|^{x\ne\dfrac{1}{3}}_{x\ne3}\)
<=>9x2-6x+1-(x2-6x+9)=0
<=>9x2-6x+1-x2+6x-9=0
<=>8x2-8=0
<=>8x2=8
<=>x2=1
<=>x2=\(\pm1\)(TMĐK)
`a,(3x-1)^2-(x+3)^2=0`
`<=>(3x-1-x-3)(3x-1+x+3)=0`
`<=>(2x-4)(4x+2)=0`
`<=>(x-2)(2x+1)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=-\dfrac{1}{2}\end{array} \right.$
`b,x^3=x/49`
`<=>x(x^2-1/49)=0`
`<=>x(x-1/7)(x+1/7)=0`
`<=>` $\left[ \begin{array}{l}x=0\\x=\dfrac{1}{7}\\x=-\dfrac{1}{7}\end{array} \right.$
`c,x^2-7x+12=0`
`<=>x^2-3x-4x+12=0`
`<=>x(x-3)-4(x-3)=0`
`<=>(x-3)(x-4)=0`
`<=>` $\left[ \begin{array}{l}x=3\\x=4\end{array} \right.$
`d,4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+x-1=0`
`<=>(x-1)(4x+1)=0`
`<=>` $\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{4}\end{array} \right.$
`e,x^3-2x-4=0`
`<=>x^3-8-2x+4=0`
`<=>(x-2)(x^2+2x+4)-2(x-2)=0`
`<=>(x-2)(x^2+2x+2)=0`
`x^2+2x+2>=2>0`
`=>x=2`
`f,x^3+8x^2+17x+10=0`
`<=>x^3+x^2+7x^2+7x+10x+10=0`
`<=>x^2(x+1)+7x(x+1)+10=0`
`<=>(x+1)(x^2+7x+10)=0`
`<=>(x+1)(x^2+2x+5x+10)=0`
`<=>(x+1)[x(x+2)+5(x+2)]=0`
`<=>(x+1)(x+2)(x+5)=0`
`<=>` $\left[ \begin{array}{l}x=-2\\x=-1\\x=-5\end{array} \right.$
`g,x^3+3x^2+6x+4=0`
`<=>x^3+x^2+2x^2+2x+4x+4=0`
`<=>x^2(x+1)+2x(x+1)+4(x+1)=0`
`<=>(x+1)(x^2+2x+4)=0`
`<=>x+1=0`
`<=>x=-1`
`h,x^3-11x^2+30x=0`
`<=>x(x^2-11x+30)=0`
`<=>x(x^2-5x-6x+30)=0`
`<=>x[x(x-5)-6(x-5)]=0`
`<=>x(x-5)(x-6)=0`
`<=>` $\left[ \begin{array}{l}x=0\\x=5\\x=6\end{array} \right.$