Phân tích đa thức thành nhân tử
a) \(x^8+x^4+1\)
b) \(x^5+x^4+1\)
c) \(x^8+x+1\)
d) \(x^8+x^7+1\)
e) \(x^3-6x^2+11x-6\)
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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)
AD
23 tháng 7 2023
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
6 tháng 7 2017
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
T
13 tháng 1
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
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.
a) x8+x4+1 = (x8+x7+x6) +(-x7-x6-x5)+(x5+x4+x3)+(-x3-x2-x)+(x2+x+1) = (x2+x+1)(x6-x5+x3-x+1)
b) x5+x4+1 = x5 +x4+x3-x3-x2-x+x2+x+1=(x2+x+1)(x3-x+1)
tương tự thì c) và d) cx có nhân tử x2+x+1
e) = x3-x2-5x2+5x+6x+6 = (x-1)(x2-5x+6) = (x-1)(x2-2x-3x+6) = (x-1)(x-2)(x-3)
a) Ta có: \(x^8+x^4+1=\left(x^4\right)^2+2.x^4.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\) Không phân tích được