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\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
a, 7x - 14
= 7(x-2)
b, 2x - 2y + \(x^2\)- xy
= (2x-2y) + (\(x^2\)-xy)
= 2(x-y) + x(x-y)
= (x-y)(2+x)
c, 6x + 12
= 6(x+2)
\(a,=7\left(x-2\right)\\ b,=2\left(x-y\right)+x\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=6\left(x+2\right)\\ d,\text{Sai đề}\)
Lời giải của các bạn đều thỏa mãn yêu cầu đề bài là phân tích đa thức thành nhân tử
\(3x^4-48\)
\(=\left(3x^4-6x^3\right)+\left(6x^3-12x^2\right)+\left(12x^2-24x\right)+\left(24x-48\right)\)
\(=3x^3\left(x-2\right)+6x^2\left(x-2\right)+12x\left(x-2\right)+24\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(3x^3+6x^2\right)+\left(12x+24\right)\right]\)
\(=\left(x-2\right)\left[3x^2\left(x+2\right)+12\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(3x^2+12\right)\)
\(x^4-8x\)
\(=x\left(x^3-8\right)\)
\(=x\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(4x-8\right)\right]\)
\(=x\left[x^2\left(x-2\right)+2x\left(x-2\right)+4\left(x-2\right)\right]\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
Giải:
a) \(3x^2y-6xy^2\)
\(=3xy\left(x-2y\right)\)
Vậy ...
b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)
\(=\left(2x-a\right)\left(x^2-y\right)\)
\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)
Vậy ...
c) \(25a^2-c^2\)
\(=\left(5a-c\right)\left(5a+c\right)\)
Vậy ...
d) \(4-36x+81x^2\)
\(=2^2-2.2.9x+\left(9x\right)^2\)
\(=\left(2-9x\right)^2\)
Vậy ...
e) \(\left(x+7\right)2-\left(2x-9\right)2\)
\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)
\(=2\left(x+7-2x+9\right)\)
\(=2\left(16-x\right)\)
Vậy ...
f) \(x^2-6x+8\)
\(=x^2-6x+9-1\)
\(=\left(x-3\right)^2-1\)
\(=\left(x-4\right)\left(x-2\right)\)
Vậy ...
(x2 + 2.x.3 + 32 - 1).(x2 + 2.x.4 + 16 - 1) - 24
=[(x+3)2 - 1]. [(x+4)2-1] -24
=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24
=(x+4)(x+2)(x+5)(x-3) - 24
(x2+6x+8)(x2+8x+15)-24
<=>(x2+4x+2x+8)(x2+5x+3x+15)-24
<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24
<=> (x+4)(x+2)(x+5)(x+3)-24
<=> (x+4)(x+3)(x+2)(x+5)-24
<=>(x2+7x+12)(x2+7x+10)
đặt t=x2+7x+11 ta có:
(t-1)(t+1)-24
<=> t2-1-24
<=>t2-25
<=>(t-5)(t+5)
thay t=x2+7x+11 vào ta có:
(x2+7x+11-5)(x2+7x+11+5)
<=>(x2+7x+6)(x2+7x+16)
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)