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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
(x2 - x)2 - 2 * (x2 - x) - 15
đặt x2 - x = a
có: a2 - 2a - 15 = (a2 - 2a + 1) - 16 = (a - 1)2 - 16 = (a - 5) (a + 3)
thay vào đc: (x2 - x - 5) (x2 - x +3)
(x^2+x)^2-2(x^2+x)-15
=(x2+x)2-2(x2+x)+1-16
=(x2+x-1)2-16
=(x2+x-1+4)(x2+x-1-4)
=(x2+x+3)(x2+x-5)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
( x2 + 8x + 7 ) ( x2 + 8x + 15 ) + 15
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
=(x^2+8x)^2+23(x^2+8x)+135
Cái này ko phân tích được nha bạn
\(x^4+x^3-20x^2-47x-15\)
\(=x^3\left(x-5\right)+6x^2\left(x-5\right)+10x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x-5\right)\left(x^3+6x^2+10x+3\right)\)
\(=\left(x-5\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)+\left(x+3\right)\right]\)
\(=\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)
\(=x^4-5x^3+6x^3-30x^2+10x^2-50x+3x-15\\ =\left(x-5\right)\left(x^3+6x^2+10x+3\right)\\ =\left(x-5\right)\left(x^3+3x^2+3x^2+9x+x+3\right)\\ =\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)
đặt x^2 - x = t
phương trình trở thành
t^2-2t-15
= t^2 - 2t + 1 -16
= (t^2 - 1) - 16
=> (x^2 - x - 1) -16
đặt x^2 - x = t
phương trình trở thành
t^2-2t-15
= t^2 - 2t + 1 -16
= (t^2 - 1) - 16
=> (x^2 - x - 1) -16