a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)

\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)

\(=2^4.5+2-5^2\)

\(=57\)

b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)

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

c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)

\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)

\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4

giúp với trí ơi

giúp nốt trí ui

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

\(=\left(2x+y\right)\left(2x-y\right)-\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\)

b) \(9x^2-25y^2-6x+10y=\left(3x-5y\right)\left(3x-5y\right)-2\left(3x-5y\right)\)

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

c) \(x^3-2x^2+2x-1=x^3-1-2x^2+2x\)

\(=\left(x-1\right)\left(x^2+x+1\right)-2x\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1-2x\right)=\left(x-1\right)\left(x^2-x+1\right)\)

d) \(x^4+2x^3-4x-4=x^4-4+2x^3-4x\)

\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)=\left(x^2-2\right)\left(x^2+2+2x\right)\)

Giúp tôi với tôi sắp thi rồi

thông cảm nhé tôi cũng có 1 bài giống hệt bạn luôn

\(3.x^{4+5}.x^2.y^{2+2}.y^{4+2}.y^2\)

NTN đúng ko??