Question

[Lhp20888l]

 

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 2:

a: Đặt A(x)=0

=>\(\left(2x+3\right)\left(x-1\right)-2x\left(x+3\right)=0\)

=>\(2x^2-2x+3x-3-2x^2-6x=0\)

=>-5x-3=0

=>-5x=3

=>\(x=-\dfrac{3}{5}\)

b: Đặt B(x)=0

=>\(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

c: Đặt C(x)=0

=>\(\left(x^2+1\right)\left(x^2-3\right)=0\)

mà \(x^2+1>0\forall x\)

nên \(x^2-3=0\)

=>\(x^2=3\)

=>\(x=\pm\sqrt{3}\)

Câu 1:

a: \(A\left(x\right)=x^4-3x^3-3x^4+2x^3-x^2+2x-5\)

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

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

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

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

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

b: A(x)+B(x)

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

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

A(x)-B(x)

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

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

c: \(B\left(1\right)=2\cdot1^4-3\cdot1^3-1^2+2\cdot1=2-3-1+2=0\)