a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

a) \(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2.4=25-8=17\)

b) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.4.5=125-60=65\)

c) \(x^4+y^4=\left(x^2\right)^2+\left(y^2\right)^2=\left(x^2+y^2\right)^2-2x^2y^2\)

\(=\left(\left(x+y\right)^2-2xy\right)^2-2\left(xy\right)^2=\left(5^2-2.4\right)^2-2.4^2\)

\(=\left(25-8\right)^2-2.16=17^2-32=289-32=257\)

d) \(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)

\(=\left(x+y\right)^5-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

\(=\left(x+y\right)^5-5xy\left(\left(x^3+y^3\right)+\left(2x^2y+2xy^2\right)\right)\)

\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-3xy\left(x+y\right)+\left(2xy\left(x+y\right)\right)\right)\)

\(=\left(5\right)^5-5.4\left(\left(\left(5^3-3.4.5\right)+\left(2.4.5\right)\right)\right)\)

\(=3125-20\left(125-65+40\right)\)

\(=3125-20\left(100\right)=3125-2000=1125\)

\(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2\cdot4=25-8=17\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot4\cdot5=125-60=65\\ x^4+y^4 \\ =\left(x+y\right)^4-4xy\left(x^2+y^2\right)-6x^2y^2\\ =5^4-4\cdot4\left[\left(x+y\right)^2-2xy\right]-6\left(xy\right)^2\\ =5^4-4\cdot4\cdot\left(5^2-2\cdot4\right)-6\cdot4^2\\ =625-16\cdot\left(25-8\right)-6\cdot16\\ =625-16\cdot17-96\\ =625-272-96\\ =257\\ x^5+y^5\\ =\left(x+y\right)^5-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\\ =5^5-5\cdot4\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-10\left(xy\right)^2\cdot5\\ =3125-20\left(5^3-3\cdot4\cdot5\right)-10\cdot4^2\cdot5\\ =3125-20\cdot\left(125-60\right)-10\cdot16\cdot5\\ =3125-20\cdot65-800\\ =3125-1300-800\\ =1025\)

a: \(=\dfrac{6x+12+4-2x}{30}=\dfrac{4x+16}{30}=\dfrac{2x+8}{15}\)

b: \(=\dfrac{18x}{60}+\dfrac{8x-4}{60}+\dfrac{6-3x}{60}\)

\(=\dfrac{18x+8x-4+6-3x}{60}=\dfrac{23x+2}{60}\)

c: \(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)

d: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{x\left(y-x\right)}\)

\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}=\dfrac{x-y}{xy}\)

e: \(=\dfrac{x^2+2xy+y^2+x^2+y^2}{x+y}=\dfrac{2x^2+2xy+2y^2}{x+y}\)

Giúp mình nhé mọi người !

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=x-1\)

\(b.\)

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2y}{\left(x-y\right)}\)

Tương tự các câu còn lại

Lời giải:

\(B=x(x^2+xy+y^2)-y(y^2+xy+y^2)\)

\(=(x-y)(x^2+xy+y^2)=x^3-y^3=10^3-(-1)^3=1000-(-1)=1001\)

\(C=x^4+10x^3+10x^2+10\)

\(=x^4+9x^3+x^3+9x^2+x^2+10\)

\(=x^3(x+9)+x^2(x+9)+x^2+10\)

\(=(x+9)(x^3+x^2)+x^2+10\)

\(=(-9+9)[(-9)^3+(-9)^2]+(-9)^2+10\)

\(=0+(-9)^2+10=91\)

Thay $x=-1$ vào biểu thức:

\(D=x^2(x+y)-xy(x-y)-x(y^2+1)\)

\(=(-1)^2(x+y)-(-1)y(x-y)-(-1)(y^2+1)\)

\(=x+y+y(x-y)+(y^2+1)\)

\(=x+y+xy-y^2+y^2+1=x+y+xy+1\)

\(=(x+1)(y+1)=(-1+1)(y+1)=0\)

1,Thực hiện phép tính :

a, (x + 2)⁹ : (x + 2)⁶

=(x+2)^9-6

=(x+2)³

b, (x - y) ⁴ : (x - 2)³

=(x-y)^4-3

=x-y

c, ( x²+ 2x + 4)⁵ : (x² + 2x + 4)

=(x²+2x+4)^5-1

=(x²+2x+4)⁴

d, 2(x² + 1)³ : 1/3(x² + 1)

=(2÷1/3).[(x²+1)³÷(x²+1)]

=6(x²+1)²

e, 5 (x - y)⁵ : 5/6 (x - y)²

=(5÷5/6).[(x-y)⁵÷(x-y)²]

=6(x-y)⁾³