Phân tích các đa thức sau thành nhân tử:
1, \(x^2-2021+2020\)
2, \(x^4+64\)
3, \(x^4+x^2+1\)
4, \(x^5+x+1\)
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24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
N
1
17 tháng 12 2017
1)7x(x-5)-x(x-5)=(x-5)(7x-x)=6x(x-5)
2)x4+3x3+x+3=x3(x+3)+(x+3)=(x+3)(x3+1)=(x+3)(x+1)(x2-x+1)
3)x4+64=[(x2)2+2.x2.8+64]-16x2=(x2+8)2-(4x)2=(x2+4x+8)(x2-4x+8)
5 tháng 10 2021
Bài 1:
a: \(5x^3+10xy=5x\left(x^2+2y\right)\)
b: \(x^2+14x+49-y^2\)
\(=\left(x+7\right)^2-y^2\)
\(=\left(x+7+y\right)\left(x+7-y\right)\)
17 tháng 7 2018
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
23 tháng 11 2021
\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)
23 tháng 11 2021
a) 2x2+6x=2x(x+3)
b) x4+3x3+x+3=(x4+x)+(3x3+3)=x(x3+1)+3(x3+1)=(x+3)(x3+1)
c) 64-x2-y2+2xy=-(x2-2xy+y2)+82=8-(x+y)2=(8+x+y)(8-x-y)
A= (x+5)(x+1)+(x-2)(x2+2xx+4)-(x2+x-2)
A= x2+6x+5+x3-8-x2-x+2
A= x3+(x2-x2)+(6x-x)+(5-8+2)
A= x3+5x-1
25 tháng 7 2016
\(\left(3x+1\right)^2-4\left(x-2\right)^2=9x^2+6x+1-4\left(x^2-4x+4\right)=9x^2+6x+1-4x^2+16x-16=5x^2+22x-15=\)
\(\left(5x-3\right)\left(x+5\right)\)
25 tháng 7 2016
\(9\left(2x+3\right)^2-4\left(x+1\right)^2=9\left(4x^2+12x+9\right)-4\left(x^2+2x+1\right)=36x^2+108x+81-4x^2-8x-4=32x^2+100x+77\)
\(\left(8x+11\right)\left(4x+7\right)\)
1 tháng 9 2018
Cái này chưa học bt làm mấy câu
b. x^2 + 2x - 3
= x^2 + 3x - x - 3
= x ( x - 1 ) + 3 ( x - 1 )
= ( x + 3 ) ( x - 1 )
TN
1 tháng 9 2018
\(4x^2-3x-4\)
\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)
\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)
\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)
\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)
\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)
\(x^2+2x-3\)
\(=x^2-x+3x-3\)
\(=x\left(x-1\right)+3\left(x-1\right)\)
\(=\)\(\left(x+3\right)\left(x-1\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)
đặt \(x^2+5x+5=t\)
\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
hay \(\left(1\right)=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
học tốt
13 tháng 8 2019
\(a,x^2+9x+20=x^2+4x+5x+20.\)
\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(c,x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2-2\right)-\left(2x\right)^2=\left(x^2-2x-2\right)\left(x^2+2x-2\right)\)
\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
SN
2
S
29 tháng 5 2021
x4 + x3 + 2x2 + x + 1
= (x4 + 2x2 + 1) + (x3 + x)
= (x2 + 1)2 + x (x2 + 1)
= (x2 + 1) ( x2 + 1 + x)
= (x2 + 1) (x + 1)2
1) = \(x^2-1=\left(x-1\right)\left(x+1\right)\)
2) \(=\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
3)
\(=x^4-x+x^2+x+1=x\left(x^3-1\right)+x^2+x+1=x\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(=x^5-x^2+x^2+x+1=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
1.\(x^2-2021+2020=x^2-1=\left(x+1\right)\left(x-1\right)\)
2. \(x^4+64=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
3. \(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2+x+1\right)\)
4. \(x^5+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)