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a) \(\left(x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=25\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)
b) \(x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
c) \(x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Rightarrow x=-1\)
d) \(x^3+10x^2+25x=0\)
\(\Leftrightarrow x\left(x+5\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
a) ( x2 - 1 )( x2 - 25 ) = 0
<=> \(\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)
b) x2 - 8x + 16 = 0
<=> ( x - 4 )2 = 0
<=> x - 4 = 0
<=> x = 4
c) x3 + 3x2 + 3x + 1 = 0
<=> ( x + 1 )3 = 0
<=> x + 1 = 0
<=> x = -1
d) x3 + 10x2 + 25x = 0
<=> x( x2 + 10x + 25 ) = 0
<=> x( x + 5 )2 = 0
<=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
a) \(3x^3-12x=0\)
=> \(3x\left(x^2-4\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
=> \(x^2\left(x-3\right)+\left(-4x+12\right)=0\)
=> \(x^2\left(x-3\right)-4x+12=0\)
=> \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
=> \(\left(x-3\right)\left(x^2-4\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
=> \(\left[3x-1-\left(2x-3\right)\right]\left(3x-1+2x-3\right)=0\)
=> \(\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\)
=> \(\left(x+2\right)\left(5x-4\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{5}\end{cases}}\)
d) \(x^2-4x-21=0\)
=> \(x^2+3x-7x-21=0\)
=> \(x\left(x+3\right)-7\left(x+3\right)=0\)
=> (x + 3)(x - 7) = 0 => x = -3 hoặc x = 7
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (x + 1)(3x - 10) = 0
=> x = -1 hoặc x = 10/3
a) \(3x^3-12x=0\)
\(\Leftrightarrow3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2;0;2\right\}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\in\left\{-2;2;3\right\}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x\in\left\{-2;\frac{4}{5}\right\}\)
Ta có : 3x3 - 12x = 0
=> 3x(x2 - 4) = 0
=> x(x - 2)(x + 2) = 0
=> \(x\in\left\{0;2;-2\right\}\)
b) x2(x - 3) + 12 - 4x = 0
=> x2(x - 3) - 4(x - 3) = 0
=> (x2 - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}\)
Vậy \(x\in\left\{-2;2;3\right\}\)
c) (3x - 1)2 - (2x - 3)2 = 0
=> (3x - 1 - 2x + 3)(3x - 1 + 2x - 3) = 0
=> (x + 2)(5x - 4) = 0
=> \(\orbr{\begin{cases}x+2=0\\5x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0,8\end{cases}}\)
Vậy \(x\in\left\{-2;0,8\right\}\)
d) x2 - 4x - 21 = 0
=> x2 - 7x + 3x - 21 = 0
=> x(x - 7) + 3(x - 7) = 0
=> (x + 3)(x - 7) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=7\end{cases}}\)
Vậy \(x\in\left\{-3;7\right\}\)
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (3x - 10)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}\)
Vậy \(x\in\left\{\frac{10}{3};-1\right\}\)
\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow\left(2x^3+7x^2+7x\right)+2=0\)
\(\Leftrightarrow x\left(2x^2+7x+7+2\right)=0\)
\(\Leftrightarrow x\left(2x^2+7x+9\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x+3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2+6x\right)+\left(3x+9\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{3}{2}\end{matrix}\right.\)
chúc bạn học tốt!
a, 4x2 + 4x + 1 = 25
⇔ ( 2x + 1 ) \(^2\) - 25 = 0
⇔( 2x + 1 - 5 ) ( 2x + 1 + 5 ) =0
⇔ ( 2x - 4 ) ( 2x + 6 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)
\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)
\(\Leftrightarrow-10x^2-10x=0\)
\(\Leftrightarrow-10x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c: Ta có: \(x^3+3x^2+3x+28=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
hay x=-4