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\(\left(y-2\right)\left(y-3\right)+\left(y-2\right)-1=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)
\(\Leftrightarrow\left(y-3\right)^2=0\)
\(\Leftrightarrow y=3\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)
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\(a)\)\(x^3-x^2-x+1=0\)
\(\Leftrightarrow\)\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt ~
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a) x3- x2 - x +1 = 0. ⇒ ( x3 - x2 ) - ( x - 1 ) = 0
⇒ x2. ( x - 1) - 1.( x - 1 ) = 0 ⇒ ( x2 - 1 ).(x - 1) = 0
⇒ x2 - 1 = 0 hoặc x - 1 = 0 ⇒ x2 = 1 hoặc x = 1
Vậy x = 1
b: Sửa đề: \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)
=><\(4x^2-12x+9-4x^2+9=0\)
=>-12x+18=0
=>x=3/2
c: \(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
=>(x^2-3)(x^2+2x+3)=0
=>x^2-3=0
hay \(x=\pm\sqrt{3}\)
d: =>(x+5)(2-x)=0
=>x=2 hoặc x=-5
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9x2 - 4 - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0
<=> ( 3x - 2 )( 2x - 3 ) = 0
<=> \(\orbr{\begin{cases}3x-2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)
x3 + 64 + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 + 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 2x + 13 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x2 - 2x + 13 = ( x2 - 2x + 1 ) + 12 = ( x - 1 )2 + 12 ≥ 12 > 0 ∀ x )
( x - 3 )( x2 + 4x + 9 ) + 2( x2 - 9 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 + 2x + 6 - 10 ) = 0
<=> ( x - 3 )( x2 + 6x + 5 ) = 0
<=> ( x - 3 )( x + 1 )( x + 5 ) = 0
<=> x = 3 hoặc x = -1 hoặc x = -5
<=> ( x - 3 )(
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\(2x^3-50x=0\)
<=> \(2x\left(x^2-25\right)=0\)
<=> \(2x\left(x-5\right)\left(x+5\right)=0\)
đến đây
bạn tự giải nhé
hk tốt
Bài này áp dụng hằng đẳng thức là ra mà =))
\(\left(x+3\right)^2-x^2+9=0\)
\(\Leftrightarrow x^2+6x+9-x^2=-9\)
\(\Leftrightarrow\left(x^2-x^2\right)+6x+9=-9\Leftrightarrow6x=-18\Leftrightarrow x=-3\)