Bài 5:
\(a.A=\dfrac{2020^3+1}{2020^2-2019}\\ =\dfrac{\left(2020+1\right)\left(2020^2-2020\cdot1+1\right)}{2020^2-2019}\\ =\dfrac{2021\cdot\left(2020^2-2019\right)}{2020^2-2019}\\ =2021\\ b.B=\dfrac{2020^3-1}{2020^2+2021}\\ =\dfrac{\left(2020-1\right)\left(2020^2+2020\cdot1+1\right)}{2020^2+2021}\\ =\dfrac{2019\cdot\left(2020^2+2021\right)}{2020^2+2021}\\ =2019\)
Bài 6:
a: \(A=\left(x-5\right)\left(x^2+5x+25\right)-x^3+2\)
\(=x^3-5^3-x^3+2\)
=-125+2=-123
b: \(B=\left(2x+3\right)\left(4x^2-6x+9\right)-8x\left(x^2+2\right)+16x+5\)
\(=8x^3+27-8x^3-16x+16x+5\)
=27+5=32
Bài 7:
a: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
=>\(x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
=>\(-9x^2+27x+9x^2+18x+9=15\)
=>45x=6
=>\(x=\dfrac{6}{45}=\dfrac{2}{15}\)
b: \(x\left(x-5\right)\left(x+5\right)-\left(x-2\right)\left(x^2+2x+4\right)=-17\)
=>\(x\left(x^2-25\right)-\left(x^3-8\right)=-17\)
=>\(x^3-25x-x^3+8=-17\)
=>-25x+8=-17
=>-25x=-25
=>x=1
