Vì đa thức Q(x) có nghiệm x = -1 nên Q(-1) = 0 hay

\(5.\left(-1\right)^2-5+a^2-a=0\)

\(\Leftrightarrow a^2-a=0\Leftrightarrow a\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\\a=1\end{matrix}\right.\)

Vậy a = 0 hoặc a = 1

Lời giải:

Để $Q(x)$ có nghiệm $x=-1$ thì $Q(-1)=0$

hay $5(-1)^2-5+a^2+a(-1)=0$

hay $a^2-a=0$

hay $a(a-1)=0$

$\Rightarrow a=0$ hoặc $a=1$

\(Q(x)\) có nghiệm x=-1

\(\Rightarrow Q(-1)=0\)

\(\Leftrightarrow 5.(-1)^2-5+a^2-a=0 \Leftrightarrow a^2-a=0 \Leftrightarrow \left[\begin{array}{} a=0\\ a=1 \end{array} \right.\)

\(a\in\left\{0;1\right\}\)

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$

a) \(ĐKXĐ:\hept{\begin{cases}x\ne3\\x\ne\pm2\end{cases}}\)

b) \(D=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right)\div\left(\frac{x-3}{2-x}\right)\)

\(\Leftrightarrow D=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2-x}{x-3}\)

\(\Leftrightarrow D=\frac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2+x\right)\left(x-3\right)}\)

\(\Leftrightarrow D=\frac{4x^2+8x}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow D=\frac{4x}{x-3}\)

c) Để D = 0

\(\Leftrightarrow\frac{4x}{x-3}=0\)

\(\Leftrightarrow4x=0\)

\(\Leftrightarrow x=0\)

Vậy để D = 0 \(\Leftrightarrow\)x = 0

d) Khi \(\left|2x-1\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=5\\1-2x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=-2\left(ktm\right)\end{cases}}\)

Vậy khi \(\left|2x-1\right|=5\Leftrightarrow D\in\varnothing\)

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