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\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
a) (x-y)2-(x2-2xy)
=y2-2xy+x2-x2+2xy
=y2-(-2xy+2xy)+(x2-x2)
=y2
b)(x-y)2+x2+2xy-(x+y)2
=y2-2xy+x2+x2+2xy-y2-2xy-x2
=(y2-y2)-(2xy+2xy-2xy)+(x2+x2-x2)
=x2-2xy
\(a.2x\left(x-1\right)-3\left(x^2+4x\right)+x\left(x+2\right)\)
\(=2x^2-2x-3x^2-12x+x^2+2x\)
\(=-12x\)
\(b.\left(2x-3\right)\left(3x+5\right)-\left(x-1\right)\left(6x+2\right)+3-5x\)
\(=6x+10x-9x^2-15-6x^2-2x-6x-2+3-5x\)
\(=-15x^2+3x-14\)
\(c.\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-y^2\right)\)
\(=x^3-y^3-x^3+y^3+x^2y-y^3\)
\(=y^3+x^2y\)
Bài 2:
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
=>-13x=26
hay x=-2
b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)
c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
hay \(x\in\left\{-5;2\right\}\)
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: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
a) 3x3+5+x3+x2+x-x2-x-1-4x+x+1-(4x3+4x)-6
=..................................................-4x3-4x-6
=(3x3+x3-4x3)+(5+1-6-1)+(x2-x2)+(x-x+x-4x-4x)
=0-1+0-7x
=-1-7x
b)x2+xy-xy-y2-x2+3
=(x2-x2)+(xy-xy)-y2+3
=-y2+3
viết đề chán thật
a) 3x3 + 5 + (x - 1)(x2 + x + 1) - (4x + x + 1) - 4x(x2 + 1) - 6
= 3x3 + 5 + (x - 1)(x2 + x + 1) - [(4x + x) + 1)] - 4x(x2 + 1) - 6
= 3x3 + 5 + (x - 1)(x2 + x + 1) - (5x + 1) - 4x(x2 + 1) - 6
= 3x3 + 5 + (x - 1)(x2 + x + 1) - 5x - 1 - 4x(x2 + 1) - 6
= 3x2 + (5 - 1 - 6) + (x - 1)(x2 + x + 1) - 5x - 4(x2 + 1)
= 3x2 - 2 + x(x2 + x + 1) - 1(x2 + x + 1) - 5x - 4(x2 + 1)
= 3x2 - 2 + x3 + x2 + x - x2 - x - 1 - 5x - 4x2 - 4
= -3 - 9x
b) (x - y)(x + y) - x2 + 3
= x2 - y2 - x2 + 3
= -y2 + 3
= 3 - y2