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a) \(8x^2+30x+7=0\)
\(\Rightarrow8x^2+2x+28x+7=0\)
\(\Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(4x+1\right)=0\)
\(\Rightarrow\)\(2x+7=0\) hoặc \(4x+1=0\)
\(\Rightarrow\)\(2x=-7\) ; \(4x=-1\)
\(\Rightarrow\)\(x=\frac{-7}{2}\) ; \(x=\frac{-1}{4}\)
Vậy \(x\in\left\{\frac{-7}{2};\frac{-1}{4}\right\}\)
b) \(x^3-11x^2+30x=0\)
\(\Rightarrow x\left(x^2-11x+30\right)=0\)
\(\Rightarrow x\left(x^2-6x-5x+30\right)=0\)
\(\Rightarrow x\left[x\left(x-6\right)-5\left(x-6\right)\right]=0\)
\(\Rightarrow x\left(x-5\right)\left(x-6\right)=0\)
\(\Rightarrow\)\(x=0\) hoặc \(x-5=0\) hoặc \(x-6=0\)
\(\Rightarrow\)\(x=0\) ; \(x=5\) ; \(x=6\)
Vậy \(x\in\left\{0;5;6\right\}\)
a)\(8x^2+30x+7=0\Leftrightarrow8x^2+2x+28x+7=0\Leftrightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Leftrightarrow\left(2x+7\right)\left(4x+1\right)=0\Leftrightarrow\orbr{\begin{cases}2x+7=0\\4x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
b)\(x^3-11x^2+30x=0\Leftrightarrow x\left(x^2-11x+30\right)=0\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)
\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\Leftrightarrow x\left(x-6\right)\left(x-5\right)=0\)
<=>x=0 hoặc x-6=0 hoặc x-5=0 <=> x=0 hoặc x=6 hoặc x=5
6x3 - 11x2 - x - 2
= 6x3 - 12x2 + x2 - 2x + x - 2
= ( 6x3 - 12x2 ) + ( x2 - 2x ) + ( x - 2 )
= 6x2( x - 2 ) + x( x - 2 ) + 1( x - 2 )
= ( x - 2 )( 6x2 + x + 1 )
x3+6x2+11x+6 = x3+6x2+12x-x+8-2 = (x3+6x2+12x+8) - (x+2) = (x+2)3 - (x+2) = (x+2)[(x+2)2 - 1] = (x+2)(x+2-1)(x+2+1) = (x+2)(x+1)(x+3)
Ta có:\(x^3+9x^2+11x-21\)
\(=x^3-x^2+10x^2-10x+21x-21=x^2\left(x-1\right)+10x\left(x-1\right)+21\left(x-1\right)\)
\(=\left(x^2+10x+21\right)\left(x-1\right)=\left(x^2+3x+7x+21\right)\left(x-1\right)\)
\(=\left[x\left(x+3\right)+7\left(x+3\right)\right]\left(x-1\right)\)
\(=\left(x+3\right)\left(x+7\right)\left(x-1\right)\)
x^3+9x^2+11x-21=x^3-x^2+10x^2-10x+21x-21=(x^3-x^2)+(10x^2-10x)+(21x-21)
=x^2(x-1)+10x(x-1)+21(x-1)=(x-1)(x^2+10x+21)=(x-1)(x^2+3x+7x+21)=(x-1)[(x^2+3x)+(7x+21)]
=(x-1)(x+7)(x+3)
x8 + x4 + 1
= x8 + 2x4 + 1 - x4
= [(x4)2 + 2x4 + 1] - x4
= (x4 + 1)2 - (x2)2
= ( x4 - x2 + 1 ) ( x4 + x2 + 1 )
x8+x+1
=(x8−x2)+(x2+x+1)
=x2(x6−1)+(x2+x+1)
=x2(x2+1)(x3−1)+(x2+x+1)
=x2(x3+1)(x−1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x2(x3+1)(x−1)+1]
=(x2+x+1)[x2(x4−x3+x−1)+1]
=(x2+x+1)(x6−x5+x3−x2+1)
a)\(x^2-5x+4\)
\(=x^2-x-4x+4\)
\(=x\left(x-1\right)-4\left(x-1\right)\)
=\(\left(x-1\right)\left(x-4\right)\)
b)\(4x^2-4x-3\)
\(=4x^2+2x-6x-3\)
\(=2x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x-3\right)\left(2x+1\right)\)
a) \(x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=\left(x^2-4x+4\right)-x\)
\(=\left(x-2\right)^2-x\)
\(=\left(x-2\right)^2-\left(\sqrt{x}\right)^2\)
\(=\left(x-2-\sqrt{x}\right)\left(x-2+\sqrt{x}\right)\)
b) \(4x^2-4x-3\)
\(=4x^2-4x+1-4\)
\(=\left(2x+1\right)^2-2^2\)
\(=\left(2x+1-2\right)\left(2x+1+2\right)\)
\(=\left(2x-1\right)\left(2x+3\right)\)
\(x^3+6x^2+11x+6=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
x3 - 11x2 + 30x = x(x2 - 11x + 30) = x(x2 - 6x - 5x + 30) = x[x(x - 6) - 5(x - 6)] = x(x - 5)(x - 6)
Trả lời:
x3 - 11x2 + 30x
= x ( x2 - 11x + 30 )
= x ( x2 - 5x - 6x + 30 )
= x [ ( x2 - 5x ) - ( 6x - 30 ) ]
= x [ x ( x - 5 ) - 6 ( x - 5 ) ]
= x ( x - 6 ) ( x - 5 )