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\(\dfrac{x+10}{x-2}+\dfrac{x-18}{x+2}+\dfrac{x+2}{x^2-4}=\dfrac{\left(x+10\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-18\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+12x+20+x^2-16x-36+x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-3x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x^2+4x\right)-\left(7x+14\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x\left(x+2\right)-7\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x-7\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-7}{x-2}\)
a) \(3y^2\left(2y-1\right)+y-y\left(1-y+y^2\right)-y^2+y \)
= \(6y^3-3y^2+y-y+y^2-y^3-y^2+y\)
= \(5y^3-3y^2+y\)
b)\(25x-4\left(3x-1\right)+\left(5-2x\right)7\)
= \(25x-12x+4+35-14x\)
= \(-x+39\)
c) \(11x-2\left(10x-1\right)-\left(4x-1\right)\left(-2\right)\)
= \(11x-\left(20x-2\right)-\left(-8x+2\right)\)
= \(11x-20x+2+8x-2\)
= \(-x\)
d) \(\left(\frac{1}{2x}\right)3-x\left(1-2x-\frac{1}{8x^2}\right)-x\left(x+\frac{1}{2}\right)\)
= \(\frac{3}{2x}-x+2x^2+\frac{x}{8x^2}-x^2-\frac{x}{2}\)
= \(\left(\frac{3}{2x}+\frac{1}{8x}-\frac{x}{2}\right)+x^2-x\)
= \(\left(\frac{12+1-4x^2}{8x}\right)+x^2-x\)
= \(\frac{13-4x^2}{8x}+\frac{8x^3}{8x}-\frac{8x^2}{8x}\)
= \(\frac{13-4x^2+8x^3-8x^2}{8x}\)
= \(\frac{8x^3-12x^2+13}{8x}\)
= x2 - \(\frac{3}{2}\)+\(\frac{13}{8x}\)
e) \(12\left(2-3x\right)+35x-\left(x+1\right)\left(-5\right)\)
= \(24-36x+35x-\left(-5x-5\right)\)
= \(24-36x+35x+5x+5\)
= 4x + 29
Bài2: phân tích đa thức thành nhân tử
\(a,x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(y+x-2\right)\)
\(b,x^3-5x^2+x-5\)
\(=x^2\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+x-5\right)\left(x-x-5\right)\)
\(c,x^2-2xy+y^2-9\)
\(=\left(x^2-y^2\right)-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
chúc bạn học tốt !
a) A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
A = 6x^2 + 33x - 10x - 55 - 6x^2 - 23x - 21
A = -76
b) B = 4x(3x - 2) - 3x(4x + 1)
B = 12x^2 - 8x - 12x^2 - 3x
B = -11x
c) C = (x + 3)(x - 2) - (x - 1)^2
C = x^2 + x - 6 - x^2 + 2x - 1
C = 3x - 7
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a, \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\text{[}x\left(x+1\right)+2\left(x+1\right)\text{]}\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
b, \(2x^3+3x^2+3x+2\)
\(=2x^3+2x^2+x^2+x+2x+2\)
\(=2x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+x+2\right)\)
c, \(x^3-4x^2-8x+8\)
\(=x^3+2x^2-6x^2-12x+4x+8\)
\(=x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-6x+4\right)\)
\(\dfrac{3-3x}{2x}+\dfrac{3x-1}{2x-1}+\dfrac{11x-5}{2x-4x^2}\\ =\dfrac{\left(3-3x\right)\left(1-2x\right)}{2x\left(1-2x\right)}-\dfrac{2x\left(3x-1\right)}{2x\left(1-2x\right)}+\dfrac{11x-5}{2x\left(1-2x\right)}\\ =\dfrac{3-9x+6x^2}{2x\left(1-2x\right)}-\dfrac{6x^2-2x}{2x\left(1-2x\right)}+\dfrac{11x-5}{2x\left(1-2x\right)}\\ =\dfrac{3-9x+6x^2-6x^2+2x+11x-5}{2x\left(1-2x\right)}\\ =\dfrac{-2}{2x\left(1-2x\right)}\\ =\dfrac{-1}{x\left(1-2x\right)}\)