Đáp án cần chọn là: D

\(P\left(1\right)=4-5+1-2=-2\)

\(P\left(-1\right)=-4-5-1-2=-12\)

\(P\left(-2\right)=4\cdot\left(-8\right)-5\cdot4-2-2=-56\)

\(P\left(-\dfrac{3}{2}\right)=4\cdot\dfrac{-27}{8}-5\cdot\dfrac{9}{4}+\dfrac{-3}{2}-2\)

\(=\dfrac{-27}{2}-\dfrac{45}{4}+\dfrac{-3}{2}-2=-\dfrac{113}{4}\)

Sửa đề: \(E=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)

-x+y-z=0

=>x-y+z=0

=>x+z=y; x-y=-z; -y+z=-x

=>x+z=y; x-y=-z; y-z=x

Ta có: \(E=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)

\(=\frac{x+z}{z}\cdot\frac{x-y}{x}\cdot\frac{y-z}{y}\)

\(=\frac{y}{z}\cdot\frac{-z}{x}\cdot\frac{x}{y}=-1\)

3r3reR

Lời giải:

a) $(x^2+1)(x-1)=0\Rightarrow x^2+1=0$ hoặc $x-1=0$

$\Rightarrow x^2=-1$ hoặc $x=1$. Dễ thấy TH $x^2=-1< 0$ vô lý nên $x=1$

Thay vào biểu thức $E$ thì:

$E=3+8-1=10$

b) $x-5=0$ nên:

$G=x^{2020}(x-5)+2=x^{2020}.0+2=2$

B   = x x 3 + 1 + 1 − x x 2 − x + 1 + 1 x + 1 = x (x + 1)(x 2 − x + 1) + (1 − x)(x + 1) (x + 1)(x 2 − x + 1) + 1.(x 2 − x + 1) (x + 1)(x 2 − x + 1) = x + 1 − x 2 + x 2 − x + 1 (x + 1)(x 2 − x + 1) = 2 x 3 + 1

Bài 1:

a: \(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2-12x+9\right)-5+20x\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2-12x+9\right)+20x-5\)

=12x-21-\(4x^2+12x-9\)

\(=-4x^2+24x-30\)

b: \(B=-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)

\(=-x\left(x^2+4x+4\right)+4x^2+4x+1+x^3+27-1\)

\(=-x^3-4x^2-4x+x^3+4x^2+4x+27=27\)

Bài 2:

a: \(27\left(1-x\right)\left(x^2+x+1\right)+81\left(x-1\right)\)

\(=27\left(1-x^3\right)+81x-81=27-27x^3+81x-81=-27x^3+81x-54\)