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)

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

Nghiệm của đa thức là: 1; 5/3

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

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

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

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

a: \(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{9}=\dfrac{11}{18}\)

hay \(x=\dfrac{11}{18}:\dfrac{1}{4}=\dfrac{11}{18}\cdot4=\dfrac{44}{18}=\dfrac{22}{9}\)

d: =>x+1;x-2 khác dấu

Trường hợp 1: \(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Leftrightarrow-1< x< 2\)

Trường hợp 2: \(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Leftrightarrow2< x< -1\left(loại\right)\)

e: =>x-2>0 hoặc x+2/3<0

=>x>2 hoặc x<-2/3

a)

Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)

\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)

Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :

\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)

b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:

\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)

c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)

\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)

d) \((2x-1)^3=-27=(-3)^3\)

\(\Rightarrow 2x-1=-3\)

\(\Rightarrow 2x=-2\Rightarrow x=-1\)

e) \((x-2)^2=1=1^2=(-1)^2\)

\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)

f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)

\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)

g) \((x-1)^2=(x-1)^6\)

\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)

\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)

\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)

Vậy \(x=\left\{0;1;2\right\}\)

\(C\left(x\right)=\left(x-1\right)\left(x-1-\frac{2}{3}\right)=0\)

\(=\left(x-1\right)\left(x-\frac{5}{3}\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{5}{3}\end{matrix}\right.\)