1) Rút gọn
x2 + 4x +4 \ 3x + 6 ; 12x + 3x \ 2x(x + 4)
2) Quy đồng mẫu
a/ 6 \ x2 + 4x ; 3 \ 2x + 8
b/ 1 \ x + 2 ; 1 \ (x + 2)(4x +7)
c/ y - 12 \ 6y - 36 ; 6 \ y2 - 6y
d/ 1 \ x + 3 ; 1 \ (x + 3)(x + 2) ; 1 \ (x + 2)(4x + 7)
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\(y=\frac{\frac{^x}{x^2}-x-6-x-\frac{1}{3}x^2-4x-15}{x^4}-2x^2+\frac{1}{3}x^2+11x+10b\)
\(y=\frac{-\left(5x^7-33x^6-30bx^5+x^3+18x^2+63x-3\right)}{3x^5}\)
a, \(A=-x-2,5-\left|4x+4,8\right|\)
\(b,B=-\left|-4-2,5x\right|-\dfrac{6}{5}x-6,2\)
\(c,C=\left|4,2-1,4x\right|-\dfrac{4}{5}x-\dfrac{24}{5}\)
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28
b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10
c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x
d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1
\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)
\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)
Ta có: P=|3x-6|-3x+5
=3x-6-3x+5
=-1
Ta có: Q=|8-2x|+3x+8
=2x-8+3x+8
=5x
\(a,\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\\ b,\dfrac{x^2+3xy}{x^2-9y^2}=\dfrac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\dfrac{x}{x-3y}\\ c,\dfrac{x^2+4x+4}{3x+6}=\dfrac{\left(x+2\right)^2}{3\left(x+2\right)}=\dfrac{x+2}{3}\)
1, a,= (x+2)^2/3.(x+2) = x+2/3
b, = 3x.(x+4)/2x.(x+4) = 3/2
k mk nha