phân tích đa thức thành nhân tử
a) \(x^4+4\)
b) \(4x^8+1\)
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6 tháng 12 2021
\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)
TV
3
AT
17 tháng 7 2021
a) \(3x-1=\left(\sqrt{3x}\right)^2-1^2=\left(\sqrt{3x}-1\right)\left(\sqrt{3x}+1\right)\)
b) \(4x-25=\left(2\sqrt{x}\right)^2-5^2=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)
c) \(x-3\sqrt{x}-4\left(x\ge0\right)\Rightarrow x+\sqrt{x}-4\sqrt{x}-4\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\)
HT
3
14 tháng 12 2022
a: =3x(y-2)-5(y-2)
=(y-2)(3x-5)
b: =(2x-5y)(2x+5y)
c: =x(x^3-8)
=x(x-2)(x^2+2x+4)
NH
14 tháng 12 2022
a)
\(3x\left(y-2\right)+5\left(2-y\right)\\=3x\left(y-2\right)-5\left(y-2\right)\\ =\left(y-2\right)\left(3x-5\right) \)
b)
\(4x^2-25y^2\\ =\left(2x-5y\right)\left(2x+5y\right)\)
c)
\(x^4-8x\\ =x\left(x^3-8\right)\\ =x\left(x-2\right)\left(x^2+2x+4\right)\)
KV
10 tháng 12 2023
a) x² + 6x + 8
= x² + 2x + 4x + 8
= (x² + 2x) + (4x + 8)
= x(x + 2) + 4(x + 8)
= (x + 2)(x + 4)
b) 3x² - 2(x - y)² - 3y²
= (3x² - 3y²) - 2(x - y)²
= 3(x² - y²) - 2(x - y)²
= 3(x + y)(x - y) - 2(x - y)²
= (x - y)[3(x + y) - 2(x - y)]
= (x - y)(3x + 3y - 2x + 2y)
= (x - y)(x + 5y)
c) 4x² - 9y² + 4x - 6y
= (4x² - 9y²) + (4x - 6y)
= (2x - 3y)(2x + 3y) + 2(2x - 3y)
= (2x - 3y)(2x + 3y + 2)
d) x(x + 1)² + x(x - 5) - 5(x + 1)²
= [x(x + 1)² - 5(x + 1)²] + x(x - 5)
= (x + 1)²(x - 5) + x(x - 5)
= (x - 5)[(x + 1)² + x]
= (x - 5)(x² + 2x + 1 + x)
= (x - 5)(x² + 3x + 1)
e) 2xy - x² + 3y² - 4y + 1
= -x² + 2xy - y² + 4y² - 4y + 1
= -(x² - 2xy + y²) + (4y² - 4y + 1)
= -(x - y)² + (2y - 1)²
= (2y - 1)² - (x - y)²
= (2y - 1 - x + y)(2y - 1 + x - y)
= (3y - x - 1)(x + y - 1)
f) 4x¹⁶ + 81
= (2x⁸)² + 2.2x⁸.9 + 9² - 2.2x⁸.9
= (2x⁸ + 9)² - 36x⁸
= (2x⁸ + 9) - (6x⁴)²
= (2x⁸ + 9 - 6x⁴)(2x⁸ + 9 + 6x⁴)
= (2x⁸ - 6x⁴ + 9)(2x⁸ + 6x⁴ + 9)
14 tháng 9 2021
a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
b) \(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
T
13 tháng 1 2024
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
T
13 tháng 1 2024
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.
TM
1
12 tháng 11 2021
a: \(=x^2\left(x-2\right)\)
b: \(=\left(x-3\right)\left(2x-9\right)\)
TM
1
10 tháng 11 2021
\(a,=x^2\left(x-2\right)\\ b,=\left(x-3\right)\left(2x-9\right)\\ c,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
a) \(x^4+4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) \(4x^8+1\)
\(=\left(2x^4\right)^2+2\cdot2x^4\cdot1+1^2-2\cdot2x^4\cdot1\)
\(=\left(2x^4+1\right)-\left(2x^2\right)^2\)
\(=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)
a/ \(x^4+4=\left(x^2\right)^2+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-4x+2\right)\left(x^2+4x+2\right)\)
b/ \(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2+4x^4+1-4x^4=\left(2x^4+1\right)^2-4x^2=\left(2x^4-2x+1\right)\left(2x^4+2x+1\right)\)