\(C\left(x\right)=\left(x-1\right)\left(x-1\right)-\frac{2}{3}\left(x-1\right)=\left(x-1\right)\left(x-1-\frac{2}{3}\right)=\left(x-1\right)\left(x-\frac{5}{3}\right)\)

Nghiệm của đa thức là: 1; 5/3

\(A\left(x\right)=\left(x-2x^2\right)\left(15x^2+7\right)\)

\(A\left(x\right)=0\)\(\Leftrightarrow\left(x-2x^2\right)\left(15x^2+7\right)=0\)

\(\Leftrightarrow x-2x^2=0\Leftrightarrow x\left(1-2x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-2x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\2x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

Hoặc \(\Leftrightarrow15x^2+7=0\Leftrightarrow15x^2=-7\Leftrightarrow x^2=\frac{-7}{15}\)(vô lí) 

Vậy \(x=0,x=\frac{1}{2}\)là 2 nghiệm của \(A\left(x\right)\)

\(\left(x-2x^2\right)\left(15x^2+7\right)=0\)

Với \(x-2x^2=0\)

\(\Rightarrow x=2x^2\Rightarrow2x=1\)

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

Với \(15x^2+7=0\Rightarrow15x^2=-7\)

\(x^2=-\frac{7}{15}\)vô lý)

Vậy nghiệm của đa thứ A(x) là \(x=\frac{1}{2}\)

\(A\left(x\right)=\left(x-2x^2\right).\left(15x^2+7\right)=0\)

Mà \(15x^2+7>0\)

\(\Rightarrow x-2x^2=0\Rightarrow x^2=\frac{x}{2}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

Câu 1:

a) \(P\left(x\right)=x^5+7x^4-9x^3+\left(-3x^2+x^2\right)-\frac{1}{4}x\)

\(P\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\)

\(Q\left(x\right)=-x^5+5x^4-2x^3+\left(x^2+3x^2\right)-\frac{1}{4}\)

\(Q\left(x\right)=-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)

b) \(P\left(x\right)+Q\left(x\right)=\left(x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\right)+\left(-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\right)\)

\(P\left(x\right)+Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)

\(P\left(x\right)+Q\left(x\right)=\left(x^5-x^5\right)+\left(7x^4+5x^4\right)-\left(9x^3+2x^3\right)+\left(-2x^2+4x^2\right)-\frac{1}{4}x-\frac{1}{4}\)

\(P\left(x\right)+Q\left(x\right)=12x^4-11x^3+2x^2-\frac{1}{4}-\frac{1}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\right)-\left(-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\right)\)

\(P\left(x\right)-Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x+x^5-5x^4+2x^3-4x^2+\frac{1}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(x^5+x^5\right)+\left(7x^4-5x^4\right)+\left(-9x^3+2x^3\right)-\left(2x^2+4x^2\right)-\frac{1}{4}x+\frac{1}{4}\)

\(P\left(x\right)-Q\left(x\right)=2x^5+2x^4-7x^3-6x^2-\frac{1}{4}x+\frac{1}{4}\)

c) \(P\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\)

\(P\left(0\right)=0^5+7\cdot0^4-9\cdot0^3-2\cdot0^2-\frac{1}{4}\cdot0\)

\(P\left(0\right)=0\)

\(Q\left(x\right)=-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)

\(Q\left(0\right)=0^5+5\cdot0^4-2\cdot0^3+4\cdot0^2-\frac{1}{4}\)

\(Q\left(0\right)=-\frac{1}{4}\)

Vậy \(x=0\) là nghiệm của đa thức P(x) nhưng không là nghiệm của đa thức Q(x)

a) \(H\left(x\right)=3x^2+2x+2012=3\left(x^2+\frac{2}{3}x+\frac{2012}{3}\right)\)

\(=3\left(x^2+2.x.\frac{1}{3}+\frac{1}{9}-\frac{1}{9}+\frac{2012}{3}\right)\)

\(=3\left[\left(x+\frac{1}{3}\right)^2+\frac{6035}{9}\right]=3\left(x+\frac{1}{3}\right)^2+\frac{6035}{3}\ge\frac{6035}{3}>0\forall x\)

Vậy đa thức vô nghiệm

b) \(D\left(x\right)=x^2+4x+4=0\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Nghiệm của đa thức là -2

c)\(F\left(x\right)=x^3-2x^2-2x+4=0\)

\(\Leftrightarrow x^2\left(x-2\right)-2\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x^2-2=0\left(1\right)\end{cases}}\).Xét đa thức (1): \(x^2-2=0\Leftrightarrow x^2=2\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

Vậy...

a, Vô nghiệm

b, Nghiệm là x = -2

Học tốt

Đa thức \(h\left(x\right)=x^3+3x^2+3x+1.\)có nghiệm 

\(\Leftrightarrow x^3+3x^2+3x+1=0\)

\(\Leftrightarrow x^2.\left(1+3x\right)+\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right).\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x^2+1=0\left(ktm\right)\end{cases}\Rightarrow x=-\frac{1}{3}}\)

Vậy   .........

Ta có: \(h\left(x\right)=0\Leftrightarrow x^3+3x^2+3x+1=0\) 

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

                               \(\Leftrightarrow x^2.\left(x+1\right)+2x.\left(x+1\right)+\left(x+1\right)=0\)

                                \(\Leftrightarrow\left(x^2+2x+1\right).\left(x+1\right)=0\)

                                 \(\Leftrightarrow\left[\left(x^2+x\right)+\left(x+1\right)\right].\left(x+1\right)=0\)

                                  \(\Leftrightarrow\left[x.\left(x+1\right)+\left(x+1\right)\right].\left(x+1\right)=0\)

                                 \(\Leftrightarrow\left(x+1\right).\left(x+1\right).\left(x+1\right)=0\)

                                  \(\Leftrightarrow\left(x+1\right)^3=0\)

                                    \(\Leftrightarrow x+1=0\)

                                    \(\Leftrightarrow x=-1\)

Vậy...

sorry nhìn nhầm 

Đa thức h(x) có nghiệm 

<=> x³+3x²+3x+1=0

<=> x³+2x²+x²+2x+x+1=0

<=> x².(x+1)+2x.(x+1)+(x+1)=0

<=>(x+1).(x²+2x+1)=0

<=> (x+1)³=0

<=> x+1=0

<=> x=-1

Vậy............

x¹⁹⁹⁴+x¹⁹⁹³+1:x²+x+1

=(x¹⁹⁹⁴+x¹⁹⁹³:x²+x)+1

=x⁹⁹⁶+1

vậy dư là x⁹⁹⁶+1

chắc zậy

Câu 1 tự lm.

Câu 2:

Ta có: \(f\left(x\right)=x^{1994}+x^{1993}+1\)

= \(\left(x^{1994}-x^2\right)+\left(x^{1993}-x\right)+\left(x^2+x+1\right)\)

= \(x^2\left(x^{1992}-1\right)+x\left(x^{1992}-1\right)+\left(x^2+x+1\right)\)

= \(\left[\left(x^3\right)^{664}-\left(1^3\right)^{664}\right]\left(x^2+x\right)+\left(x^2+x+1\right)\)

= \(\left(x^3-1^3\right)\left(x^{1989}+x^{1986}+...+x^3+1\right)+\left(x^2+x+1\right)\)

= \(\left(x-1\right)\left(x^2+x+1\right)\left(x^{1989}+x^{1986}+..+1\right)+\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left[\left(x-1\right)\left(x^{1989}+..+1\right)+1\right]\)

Vì \(x^2+x+1\) \(⋮\) \(x^2+x+1\)

=> \(f\left(x\right)\) \(⋮\) \(x^2+x+1\) hay số dư trong phép chia là 0