Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x³y + x - y - 1
= (x³y - y) + (x - 1)
= y(x³ - 1) + (x - 1)
= y(x - 1)(x² + x + 1) + (x - 1)
= (x - 1)[y(x² + x + 1) + 1]
= (x - 1)(x²y + xy + y + 1)
b) x²(x - 2) + 4(2 - x)
= x²(x - 2) - 4(x - 2)
= (x - 2)(x² - 4)
= (x - 2)(x - 2)(x + 2)
= (x - 2)²(x + 2)
c) x³ - x² - 20x
= x(x² - x - 20)
= x(x² + 4x - 5x - 20)
= x[(x² + 4x) - (5x + 20)]
= x[x(x + 4) - 5(x + 4)]
= x(x + 4)(x - 5)
d) (x² + 1)² - (x + 1)²
= (x² + 1 - x - 1)(x² + 1 + x + 1)
= (x² - x)(x² + x + 2)
= x(x - 1)(x² + x + 2)
e) 6x² - 7x + 2
= 6x² - 3x - 4x + 2
= (6x² - 3x) - (4x - 2)
= 3x(2x - 1) - 2(2x - 1)
= (2x - 1)(3x - 2)
f) x⁴ + 8x² + 12
= x⁴ + 2x² + 6x² + 12
= (x⁴ + 2x²) + (6x² + 12)
= x²(x² + 2) + 6(x² + 2)
= (x² + 2)(x² + 6)
g) (x³ + x + 1)(x³ + x) - 2
Đặt u = x³ + x
x³ + x + 1 = u + 1
(u + 1).u - 2
= u² + u - 2
= u² - u + 2u - 2
= (u² - u) + (2u - 2)
= u(u - 1) + 2(u - 1)
= (u - 1)(u + 2)
= (x³ + x - 1)(x³ + x + 2)
= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)
= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]
= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]
= (x³ + x - 1)(x - 1)(x² - x + 2)
h) (x + 1)(x + 2)(x + 3)(x + 4) - 1
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1
= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)
Đặt u = x² + 5x + 4
u + 2 = x² + 5x + 6
(1) u.(u + 2) - 1
= u² + 2u - 1
= u² + 2u + 1 - 2
= (u² + 2u + 1) - 2
= (u + 1)² - 2
= (u + 1 + √2)(u + 1 - √2)
= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)
= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
(x - 2)(x2 + 2x + 4) + 2(x2 - 4) - 5(x - 2) = 0
(x - 2)(x + 2)2 + 2(x - 2)(x+2) - 5(x - 2) = 0
(x - 2)[(x+2)2 + 2(x+2) - 5]= 0
(x - 2)[(x + 2)2 + 2(x + 2) + 1 - 6] = 0
( x - 2)[(x + 2 + 1)2 - 6] = 0
(x - 2)[(x + 3)2 - 6] = 0
(x - 2)(x + 3 - \(\sqrt{6}\))(x + 3 + \(\sqrt{6}\)) = 0
TH1. x - 2 = 0 <=> x = 2
TH2. x + 3 - \(\sqrt{6}\) = 0 <=> x = \(\sqrt{6}-3\)
TH3. x + 3 + \(\sqrt{6}\) = 0 <=> x = \(-\sqrt{6}-3\)
S = {2; \(\sqrt{6}-3\); \(-\sqrt{6}-3\)}
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
\(x.\left(x^3y-x\right)-x^2.\left(x^2y-2\right)=4\)
\(\Rightarrow x^4y-x^2-x^4y+x^2.2=4\)
\(\Rightarrow\left(x^4y-x^4y\right)-x^2+x^2.2=4\)
\(\Rightarrow0-x^2+x^2.2=4\)
\(\Rightarrow-x^2+x^2.2=4\)
\(\Rightarrow x^2.\left(-1+2\right)=4\)
\(\Rightarrow x^2=4\)
\(\Rightarrow x=\pm2\)
Mà đề ra: \(x>0\)
Vậy \(x=2\)
\(x.\left(x^3y-x\right)-x^2.\left(x^2y-2\right)=4\)
\(\rightarrow x^4t-x^2-x^4y+2x^2=4\)
\(\rightarrow x^2=4\)
\(\rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Mà \(x>0\)
\(\rightarrow x=2\)