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#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
(x+1).(x+2).(x+3).(x+4)-4
=(x+1)(x+4)(x+2)(x+3)-4
=(x2+5x+4)(x2+5x+6)-4
Đặt t=x2+5x+4 ta được:
t.(t+2)-4
=t2+2t-4
Vẫn sai đề
(x+2).(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)-24
=(x2+7x+10)(x2+7x+12)-24
=(x2+7x+11-1)(x2+7x+11+1)-24
Đặt x2+7x+11=a thì
=(a-1)(a+1)-24
=a2-1-24=a2-25=a2-52
=(a+5)(a-5)
=(x2+7x+16)(x2+7x+6)
( x+3) (x+2) (x+4) (x+5) -24
=(x+3)(x+4)(x+2)(x+5)-24
=(x2+7x+12)(x2+7x+10)-24
Đặt t=x2+7x+10 ta được:
(t+2)t-24
=t2+2t-24
=t2+4t-6t-24
=t.(t+4)-6.(t+4)
=(t+4)(t-6)
thay t=x2+7t+10 ta được:
(x2+7x+14)(x2+7+4)
Vậy ( x+3) (x+2) (x+4) (x+5) -24=(x2+7x+14)(x2+7x+4)
\(x^2-4-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+2-3\right)=\left(x-2\right)\left(x-1\right)\)
x2 - 4 - 3(x - 2)
= (x - 2)(x + 2) - 3(x - 2)
= (x - 2)(x + 2 - 3)
= (x - 2)(x - 1)
=[(x+1)(x+4)][(x+2)(x+3)]+8=(x2+5x+4)(x2+5x+6)+8
Đặt x2+5x+4=t
Ta có : t(t+2)+8=t2+2t-8=(t-2)(t+4)
k mk nha
\(x^3-x^2-4\)
\(=x^3-8-x^2+4\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4-x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-2\right)\)