Câu 1:

b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)

\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\dfrac{2}{x+3}\)

c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)

Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)

\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)

\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)

TA CÓ

\(p\left(\frac{1}{2}\right)=4\cdot\left(\frac{1}{2}\right)^2-4\cdot\frac{1}{2}+1=4\cdot\frac{1}{4}-2+1\)

\(=1-2+1=0\)

vậy ......

TA CÓ

\(x^2\ge0\Rightarrow4x^2\ge0\Rightarrow4x^2+1\ge1\)hay\(4x^2+1>0\)

vậy..............

Thay \(x=\frac{1}{2}\)vào P (x) ta có:

\(P\left(\frac{1}{2}\right)=4.\left(\frac{1}{2}\right)^2-4.\frac{1}{2}+1\)

\(P\left(\frac{1}{2}\right)=4.\frac{1}{4}-2+1\)

\(P\left(\frac{1}{2}\right)=1-2+1\)

\(P\left(\frac{1}{2}\right)=0\)

Vậy \(x=\frac{1}{2}\) là nghiệm của P(x)

Ta có :

\(4x^2\ge0\)

\(1>0\)

\(\Rightarrow4x^2+1>0\)

=> Đa thức Q(x) vô nghiệm

\(A.x=x+x^2+x^3+...+x^{101}\)

\(A.x-A=x^{101}-1\Rightarrow A\left(x-1\right)=x^{101}-1\)

\(\Rightarrow A=\dfrac{x^{101}-1}{x-1}\)

ta có 
`((x+2)(x-1))/(x^2-1)`

`=((x+2)(x-1))/((x-1)(x+1))`

`=(x+2)/(x-1)`

`=> ((x+2)(x-1))/(x^2-1) = (x+2)/(x-1)`

\(f\left(x\right)=x\left(x+1\right)+\left(x+1\right)+1\)

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

Vì \(\left(x+1\right)^2\ge0\)

\(\Rightarrow f\left(x\right)=\left(x+1\right)^2+1\ge1>0\)

\(\Rightarrow f\left(x\right)\) vô nghiệm

Vậy đa thức \(f\left(x\right)\) không có nghiệm

*Chứng tỏ \(x=\frac{1}{2}\) là nghiệm của đa thức \(P\left(x\right)=4x^2-4x+1\)

Cho \(P\left(x\right)=0\)

\(\Rightarrow4x^2-4x+1=0\)

\(\Rightarrow4x^2-2x-2x+1=0\)

\(\Rightarrow2x\left(2x-1\right)-\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=0\)

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

\(\Rightarrow2x-1=0\)

\(\Rightarrow x=\frac{1}{2}\)

\(\Rightarrow P\left(x\right)\) có nghiệm là \(x=\frac{1}{2}\)

\(\Rightarrowđpcm\)

*Chứng tỏ đa thức \(Q\left(x\right)=4x^2+1\) không có nghiệm

Ta có: \(4x^2\ge0\forall x\)

\(\Rightarrow4x^2+1>0\)

hay \(Q\left(x\right)>0\)

\(\Rightarrow\)Đa thức \(Q\left(x\right)=4x^2+1\) không có nghiệm   (đpcm)

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

\(=-4\)