\(a)\)

\(\left(x-1\right)\left(x^3+x^2+x+1\right)\)

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

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

\(=x^4-1\)

\(b)\)

\(\left(x+1\right)\left(x^4-x^3+x^2-1x+1\right)\)

\(=x^5+1^5\)

\(=x^5+1\)

\(\left(2x-y\right)^3-3x.\left(y-2x\right)^2-4x^2+4xy-y^2\)

\(=\left[\left(2x-y\right)^3-3x.\left(y-2x\right)^2\right]-\left[4x^2+4xy-y^2\right]\)

\(=\left[\left(2x-y\right)^2.\left(2x-y\right)-3x.\left(y-2x\right)^2\right]-\left(2x-y\right)^2\)

\(=\left[\left(y-2x\right)^2.\left(2x-y-3x\right)\right]-\left(2x-y\right)^2\)

=\(\left(y-2x\right)^2.\left(-x-y\right)-\left(2x-y\right)^2\)

\(=\left(y-2x\right)^2.\left(-x-y-1\right)\)

Chúc bạn học tốt 

phân tích đa thức

(2x – y)^3 – 3x(y – 2x)2 – 4x2 + 4xy – y ^2  

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

đây nha bạn 

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

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

\(=\left(2x-y\right)^3-\left(2x-y\right)^2\left(3x+1\right)\)

\(=\left(2x-y\right)^2\left(2x-y-3x-1\right)\)

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

\(\left(x-3\right)\left(4x-1\right)\left(6x-2\right)\left(2x+1\right)\)

\(\left(4x^2-12x-x+3\right)\left(12x^2-4x+6x-2\right)\)

\(\left(4x^2-13x+3\right)\left(12x^2+2x-2\right)\)

\(48x^4-13x^3+36x^2-156x^3-26x^2+26x-8x^2+26x-6\)

\(48x^4-169x^3+2x^2+52x-6\)

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

\(\left(x-3\right)\left(4x-1\right)-\left(6x+2\right)\left(2x-1\right)\)

\(=4x^2-x-12x+3-\left(12x^2-6x+4x-2\right)\)

\(=4x^2-x-12x+3-12x^2+6x-4x+2\)

\(=-8x^2-11x+5\)

= \(\left(xy-5\right)\left(xy+5\right)-3\left(xy-5\right)\)

= \(\left(xy-5\right)\left(xy+5-3\right)\)

\(=\left(xy-5\right)\left(xy+2\right)\)

có bị chấn thương nha bạn đã nhấn chữ manh có nghĩa đá rất manh thì phải đâu nha

Thay x = -2 vào A ta được 

\(A=\frac{\left|-2+1\right|+2.\left(-2\right)}{3.\left(-2\right)^2-2\left(-2\right)-1}=\frac{1-4}{12+4-1}=\frac{-3}{15}=-\frac{1}{5}\)

Thay x = 3/4 vào A ta được : 

\(A=\frac{\left|\frac{3}{4}+1\right|+\frac{2.3}{4}}{3\left(\frac{3}{4}\right)^2-\frac{2.3}{4}-1}=\frac{\frac{7}{4}+\frac{6}{4}}{\frac{3.9}{16}-\frac{6}{4}-1}=\frac{\frac{13}{4}}{-\frac{13}{16}}=-\frac{16}{4}=-4\)