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\(x^2-8x-9\)
\(=x^2-9x+x-9\)
\(=x\left(x-9\right)+\left(x-9\right)\)
\(=\left(x-9\right)\left(x+1\right)\)
\(8x^2-2\)
\(=2.\left(1^2-4x^2\right)\)
\(=2.\left(1-4x\right).\left(1+4x\right)\)
\(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right).\left(x-3+y\right)\)
\(x^2+8x-9\)
\(=x^2-x+9x-9\)
\(=x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x+9\right)\)
Câu đầu chưa học sorry
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
\(8x\left(x^2-9\right)=0\Rightarrow8x\left(x-3\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
Mình nhầm 1 chút nhé mọi người \(x^2-8x-9\) nhé!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ta có:\(\left(x^2+8x-34\right)^2-\left(3x^2-8x+2\right)^2\)
\(=\left(x^2+8x-34+3x^2-8x+2\right)\left[x^2+8x-34-\left(3x^2-8x+2\right)\right]\)
\(=\left(4x^2-32\right)\left(x^2+8x-34-3x^2+8x-2\right)\)
\(=\left(4x^2-32\right)\left(-2x^2+16x-36\right)\)
\(=-2\left(4x^2-32\right)\left(x^2-8x+18\right)\)
cảm ơn.nhưng sao bạn không rút 4 trong \(4x^2-32\)
=> \(\left(4\left(x^2-8\right)\right)\cdot\left(-2\left(x^2-8x+18\right)\right)\)
=\(-8\left(x^2-8\right)\left(x^2-8x+18\right)\)
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
a) 3x2 - 7x + 4
= 3x2 - 3x - 4x + 4
= 3x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( 3x - 4 )
b) x2 - 6xy + 9y2 = ( x - 3y )2
c) x2 - 8x - 9
= x2 - 9x + x - 9
= x( x - 9 ) + ( x - 9 )
= ( x - 9 )( x + 1 )
a) 3x2 - 7x + 4
= 3x2 - 4x - 3x + 4
= (3x2 - 4x) - (3x - 4)
= x.(3x - 4) - (3x - 4)
= (3x - 4).(x - 1)
b) x2 - 6xy + 9y2
= x2 - 2.x.3y + (3y)2
= (x - 3y)2
c) x2 - 8x - 9
= x2 - 9x + x - 9
= (x2 - 9x) + (x - 9)
= x.(x - 9) + (x - 9)
= (x - 9).(x + 1)