phân tích đa thức thành nhân tử
a) x4+\(\dfrac{1}{4}\)y8
b) \(\dfrac{1}{4}\)x4+y8
c) 64x4+y4
d) x4+\(\dfrac{1}{64}\)y4
bài 4:phân tích mỗi đa thức sau thành nhân tích
a, 83 yz + 122yz + 6xyz + yz
b,81x4(z2 - y2) - z2 + y2
c,\(\dfrac{x^3}{8}\) - \(\dfrac{y^3}{27}\) +\(\dfrac{x}{2}\) - \(\dfrac{y}{3}\)
d, x6 + x4 + x2 y2 + y4 - y6
a, \(8^3yz+12^2yz+6xyz+yz\)
\(=512yz+144yz+6xyz+yz\)
\(=yz\left(512+14+6x+1\right)\)
\(=yz\left(527+6x\right)\)
$---$
b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)
\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)
\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)
\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)
$---$
c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)
\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)
$---$
d, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
Phân tích các đa thức sau thành nhân tử
a,x4+2x3+3x2+2x+1
b,x4-4x3+2x2+4x+1
c,x4+x3+2x2+2x+4
bài 1 : phân tích đa thức thành nhân tử
a/ 2x2 (x – 1) + 4x (1 – x) b/ x4 – 27x c/ x2 – 4x + 3 d / x4 + x2 + 1
b: \(=x\left(x-3\right)\left(x^2+3x+9\right)\)
a/ 2x^2 (x – 1) + 4x (1 – x)
= 2x^2(x – 1) – 4x (x – 1)
= (x – 1)( 2x^2 – 4x)
=2x(x – 1)(x – 2)
Bài 1: Phân tích đa thức thành nhân tử
a) x3-2x2-5x+6
b) x4+5x2+6
c) x3+4x2+5x+2
d) x4+324
Bài 1: Phân tích các đa thức sau thành nhân tử
a)x2-y2-2x+2y e)x4+4y4
b)x2(x-1)+16(1-x) f)x4-13x2+36
c)x2+4x-y2+4 g) (x2+x)2+4x2+4x-12
d)x3-3x2-3x+1 h)x6+2x5+x4-2x3-2x2+1
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
bài 1:phân tích đa thức thành nhân tử
a,x4 +5x2 +9
b,x4 + 3x2 +4
c,2x4 - x2 -1
Bài 2:tìm x biết
a,(x+1) (x+2)(x+3)(x+4)= 120
b,(x-4x+3)(x2+6x +8) +24
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
Bài 2:
\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)
Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:
\(\left(y-1\right)\left(y+1\right)=120\)
\(\Leftrightarrow y^2-1=120\)
\(\Leftrightarrow y^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)
+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)
+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)
\(\Leftrightarrow x^2+5x+16=0\)
\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
\(\Rightarrow\) loại
Vậy \(x\in\left\{1;-6\right\}\).
\(b,\) Đề thiếu vế phải rồi bạn.
phân tích đa thức thành nhân tử
a)x4 - x3y + x - y
b)a2 - 2a + 1 - 9b2
a: =x^3(x-y)+(x-y)
=(x-y)(x^3+1)
=(x-y)(x+1)(x^2-x+1)
b: =(a-1)^2-9b^2
=(a-1-3b)(a-1+3b)
Viết các đa thức sau thành tích
1. ( 3x + 2)2 - 4
2. 4x2 - 25y2
3. 4x2- 49
4. 8z3 + 27
5. \(\dfrac{9}{25}\)x4 - \(\dfrac{1}{4}\)
6. x32 - 1
7. 4x2 + 4x + 1
8. x2 - 20x + 100
9. y4 -14y2 + 49
10. 125x3 - 64y3
1. ( 3x + 2)2 - 4
= (3x+2-2)(3x+2+2)
= 3x(3x+4)
2. 4x2 - 25y2
= (2x-5y)(2x+5y)
3. 4x2- 49
=(2x-7)(2x+7)
4. 8z3 + 27
=(2z+3)(4x2-6z+9)
5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)
= \((\dfrac{3}{5}x^2-\dfrac{1}{2})(\dfrac{3}{5}x^2+\dfrac{1}{2})\)
6. x32 - 1
=(x16-1)(x16+1)
7. 4x2 + 4x + 1
=(2x+1)2
8. x2 - 20x + 100
=(x-10)2
9. y4 -14y2 + 49
=(y2-7)2
10. 125x3 - 64y3
= (5x-4y)(25x2+20xy+16y2)
1) \(\left(3x+2\right)^2-4=\left(3x+2+2\right)\left(3x+2-2\right)=3x\left(3x+4\right)\)
2) \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)
4) \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)
6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
7) \(4x^2+4x+1=\left(2x+1\right)^2\)
8) \(x^2-20x+100=\left(x-10\right)^2\)
9) \(y^4-14y^2+49=\left(y^2-7\right)^2\)
Viết các đa thức sau thành tích
1 ) (3x + 2)2 – 4
2) 4x2 – 25y2
3 ) 4x2 – 49
4) 8z3 + 27
5, \(\dfrac{9}{25}\)x4 – \(\dfrac{1}{4}\)
6, x32 – 1
Mình cần quá trình làm nữa nha, thanks
1) \(\left(3x+2\right)^2-4\\ =\left(3x+2\right)^2-2^2\\ =\left(3x+2-2\right)\left(3x+2+2\right)\\ =3x.\left(3x+4\right)\)
2) \(4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\)
3) \(4x^2-49=\left(2x\right)^2-7^2=\left(2x-7\right)\left(2x+7\right)\)
4) \(8z^3+27=\left(2z\right)^3+3^3=\left(2z+3\right)\left(4z^2+6z+9\right)\)
5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2\right)^2-\left(\dfrac{1}{2}\right)^2=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)
6) \(x^{32}-1\\ =\left(x^{16}\right)^2-1^2\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
1: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)
2: \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
3: \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)
4: \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
5: \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)