1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)
\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)
3. \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2.\left(2x-1\right)+\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)
\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Bài 1.
B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )3
= ( x2 - 4 )( x + 3 ) - ( x3 + 3x2 + 3x + 1 )
= x3 + 3x2 - 4x - 12 - x3 - 3x2 - 3x - 1
= -7x - 13
Bài 2.
64 - x2 - y2 + 2xy
= 64 - ( x2 - 2xy + y2 )
= 82 - ( x - y )2
= ( 8 - x + y )( 8 + x - y )
Bài 3.
2x3 - x2 + 2x - 1 = 0
<=> ( 2x3 - x2 ) + ( 2x - 1 ) = 0
<=> x2( 2x - 1 ) + 1( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x2 + 1 ≥ 1 > 0 ∀ x )
1) \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(B=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(B=x^2+3x^2-4x-12-x^3-3x^2-3x-1\)
\(B=-7x-13\)
2) \(64-x^2-y^2+2xy\)
\(=64-\left(x^2-2xy+y^2\right)\)
\(=64-\left(x-y\right)^2\)
\(=\left(8-x+y\right)\left(8+x-y\right)\)
3) \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2\left(2x-1\right)+2x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
\(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
\(64-x^2-y^2+2xy\)
\(=64-\left(x^2+2xy-y^2\right)\)
\(=64-\left(x^2-2xy+y^2\right)\)
\(=8^2-\left(x-y\right)^2\)
\(=\left(8-x+y\right)\left(8+x-y\right)\)
\(2x^3-x^2+2x-1=0\)
\(x^2\left(2x-1\right)+\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x-1=0\\x^2+1=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\x^2=-1\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=\varnothing\end{cases}}}\)
