Bài 2:
\(a.\left(x-1\right)^2-2\left(x-3\right)\left(x-1\right)+\left(x-3\right)^2\\ =\left[\left(x-1\right)-\left(x-3\right)\right]^2\\ =\left(x-1-x+3\right)^2\\ =2^2\\ =4\)
Vậy giá trị của bt không phụ thuộc vào biến
\(b.\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3x^2-3x\\ =\left(x^3-3x^2+3x-1\right)-\left(x^3+2^3\right)+3x^2-3x\\ =x^3-3x^2+3x-1-x^3-8+3x^2-3x\\ =-9\)
Vậy giá trị của bt không phụ thuộc vào biến
Bài 1:
a: \(\left(x-3\right)\left(x+5\right)-\left(x-2\right)\left(x+2\right)\)
\(=x^2+2x-15-\left(x^2-4\right)=x^2+2x-15-x^2+4=2x-11\)
b: \(\left(\dfrac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\dfrac{1}{16}x^3+1\right)\)
\(=\dfrac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\dfrac{1}{4}x^3+4\)
\(=\dfrac{1}{4}\left(x^3-64y^3\right)+16y^3-\dfrac{1}{4}x^3+4=4\)
c: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2=2x+15\)
d: \(\left(x-2\right)^2+\left(x-1\right)\left(x^2+x+1\right)-x\left(x-2\right)\left(x+2\right)\)
\(=x^2-4x+4+x^3-1-x\left(x^2-4\right)\)
\(=x^3+x^2-4x+3-x^3+4x=x^2+3\)
Bài 2:
a: \(\left(x-1\right)^2-2\left(x-1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(x-1-x+3\right)^2\)
\(=2^2=4\)
b: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3x^2-3x\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x\)
\(=x^3-1-x^3-8=-9\)
