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Đặt \(x^2-3x-1=a\), ta có:
\(a^2-12a+27=a^2-9a-3a+27=a\left(a-9\right)-3\left(a-9\right)=\left(a-9\right)\left(a-3\right)\)
\(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
Mà \(x^2-3x-10=x^2-5x+2x-10=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+1\right)\)
và \(x^2-3x-4=x^2+x-4x-4=x\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-4\right)\)
\(\Rightarrow\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27=\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+2\right)\)
Đặt \(x^2-3x-1=a\)thay vào biểu thức ta được :
\(a^2-12a+27\)
\(=a^2-3a-9a+27\)
\(=a\left(a-3\right)-9\left(a-3\right)\)
\(=\left(a-3\right)\left(a-9\right)\)(1)
Thay \(a=x^2-3x-1\)vào (1) ta được :
\(\left(x^2-3x-1-3\right)\left(x^2-3x-1-10\right)\)
\(=\left(x^2-3x-4\right)\left(x^2-3x-11\right)\)
Bạn Châu sai đáp án cuối
phải là (x2-3x-4)(x2-3x-10) nha
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)
\(=x^3\left(x^2-2x+1\right)-x^2\left(x^2-2x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^3-x^2\right)\)
\(=\left(x-1\right)^2x^2\left(x-1\right)=\left(x-1\right)^3x^2\)
\(=x^2\left(x^3-1\right)-3x^3\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1-3x\right)\)
\(=x^2\left(x-1\right)\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)\left(x-1\right)^2\)
\(=x^2\left(x-1\right)^3\)
Đặt x2 - 3x - 1 = k
Khi đó, ta có: A = k2 - 12k + 27 = k2 - 3x - 9x + 27 = k(k - 3) - 9(k - 3) = (k - 9)(k - 3)
=> (x2 - 3x - 1 - 9)(x2 - 3x - 1 - 3) = (x2 - 3x - 10)(x2 - 3x - 4)
= (x2 - 5x + 2x - 10)(x2 - 4x + x - 4)
= [x(x - 5) + 2(x - 5)][x(x - 4) + (x - 4)]
= (x + 2)(x - 5)(x + 1)(x - 4)
x3 - x2 - 3x + 27
= x3 + 3x2 - 4x2 - 12x + 9x + 27
= x2 ( x + 3 ) - 4x ( x + 3 ) + 9 ( x + 3 )
= ( x + 3 ) ( x2 - 4x + 9 )
x^3-x^2-3x+27
= (x+3)(x^2-4x+9)
nha bạn chúc bạn học tốt nha
\(x^3+3x^2-3x-1=\left(x^3-1\right)+\left(3x^2-3x\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left[\left(x^2+x+1\right)+3x\right]=\left(x-1\right)\left(x^2+4x+1\right)\)
x3 - x2 + 3x - 27.
= (x3 - 27) - (x2 - 3x).
= (x - 3) (x2 + 3x + 9) - x (x - 3).
= (x - 3) (x2 + 3x + 9 - x).
= (x - 3) (x2 + 2x + 9).
x3 - x2 + 3x - 27 = ( x3 - 27 ) - ( x2 - 3x )
= ( x - 3 ) ( x2 + 3x + 9 ) - x ( x - 3 )
= ( x - 3 ) ( x2 + 3x + 9 - x )
= ( x - 3 ) ( x2 - 2x + 9 )
* Chúc bạn học tốt