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Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
Bài 2:
a)x3+2x2+x
=x(x2+2x+12)
=x(x+1)2
b)xy+y2-x-y
=(xy-x)+(y2-y)
=x(y-1)+y(y-1)
=(y-1)(x+y)
Bai 1:
a) = 2x^3 + 14x^2 - 2x^3 - x^2 + 9x - 12
= 13x^2 + 9x - 12
b) = x^2 - 2x + 1 - x^2 + 4x - 4x + 16
= -2x + 17
Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó 
Có một số câu thì mình không làm được. Mong bạn thông cảm!!!
A, x2+3x+7 = x2+2.x.3/2 +(3/2)2+19/4 = (x+3/2)2 + 19/4 >=19/4
B, = (x2-7x+10)(x2-7x-10) = (x2-7x)2 - 100 >= -100
C, = 5x2+5 >=5
e)
$x^3+6x^2+12x+8=x^3+3.2.x^2+3.2^2.x+2^3=(x+2)^3$
f)
$a^3-2a^2-ab^2+2b^2=(a^3-ab^2)-(2a^2-2b^2)$
$=a(a^2-b^2)-2(a^2-b^2)=(a^2-b^2)(a-2)=(a-b)(a+b)(a-2)$
g)
$2a^2x-2a^2-2abx+4ab-2b^2=(2a^2x-2abx)-(2a^2-4ab+2b^2)$
$=2ax(a-b)-2(a-b)^2=2(a-b)(ax-a+b)$
h)
\(x^2-2xy+y^2-25=(x-y)^2-25=(x-y)^2-5^2=(x-y+5)(x-y-5)\)
a)
$4x^2-40x^4+100x^3=4x^2(1-10x^2+25x)$
b)
\(3xy(x-5)-7x+35=3xy(x-5)-7(x-5)\)
\(=(x-5)(3xy-7)\)
c)
\(a^2-am-b^2-bm=(a^2-b^2)-(am+bm)=(a-b)(a+b)-m(a+b)\)
\(=(a+b)(a-b-m)\)
d)
\(x^3-4x-x^2y+4y=(x^3-x^2y)-(4x-4y)\)
\(=x^2(x-y)-4(x-y)=(x^2-4)(x-y)=(x-2)(x+2)(x-y)\)
a) \(x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
c) \(x^2+8x-9\)
\(=x^2-x+9x-9\)
\(=x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x+9\right)\)
d) \(x^4+5x^2-6\)
\(=x^4-x^2+6x^2-6\)
\(=x^2\left(x^2-1\right)+x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+x\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)^2\)
\(\text{a) }x^2+7x+12\\ =x^2+3x+4x+12\\ =\left(x^2+3x\right)+\left(4x+12\right)\\ =x\left(x+3\right)+4\left(x+3\right)\\ =\left(x+4\right)\left(x+3\right)\\ \)
\(\text{b) }3x^2+7x-6\\ =3x^2+9x-2x-6\\ =\left(3x^2+9x\right)-\left(2x+6\right)\\ =3x\left(x+3\right)-2\left(x+3\right)\\ =\left(3x+2\right)\left(x+3\right)\\ \)
\(\text{c) }x^2+8x-9\\ =x^2+9x-x-9\\ =\left(x^2+9x\right)-\left(x+9\right)\\ =x\left(x+9\right)-\left(x+9\right)\\ =\left(x-1\right)\left(x+9\right)\\ \)
\(\text{d) }x^4+5x^2-6\\ =x^4+2x^2+3x^2-6\\ =x^4+6x^2-x^2-6\\ =\left(x^4+6x^2\right)-\left(x^2+6\right)\\ =x^2\left(x^2+6\right)-\left(x^2+6\right)\\ =\left(x^2-1\right)\left(x^2+6\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+6\right)\\ \)
a) Ta có: \(2x^4+3x^3-9x^2-3x+2\)
\(=2x^4-2x^3-2x^2+5x^3-5x^2-5x-2x^2+2x+2\)
\(=2x^2\left(x^2-x-1\right)+5x\left(x^2-x-1\right)-2\left(x^2-x-1\right)\)
\(=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)
Có A=\(\left(x^3+x^2-3x\right)+\left(-2x^2-2x+a+2\right)=-x\left(-x^2-x+3\right)-2x^2-2x+a+2⋮-x^2-x+3\)
\(\Rightarrow C=-2x^2-2x+a+2⋮B\). Chỉ có thể C=\(2\left(-x^2-x+3\right)\Rightarrow a+2=6\Rightarrow a=4\)
\(A=\left(2x^3+3x^2+4x\right)+\left(-10x^2-15x+a-8\right)=x\left(2x^2+3x+4\right)+\left(-10x^2-15x+a-8\right)⋮2x^2+3x+4\)\(\Rightarrow C=-10x^2-15x+a-8⋮2x^2+3x+4\)
Chỉ có thể C=\(-5\left(2x^2+3x+4\right)\) \(\Rightarrow a-8=-20\Rightarrow a=-12\)