\(-3\in Z\\ 0\in Z\\ 4\in Z\\ -2\notin N\)

a) \(\in\)

b) \(\in\)

c) \(\in\)

d) \(\notin\)

a) -3 ∈ Z

b) 0 ∈ Z

c) 4 ∈ Z

d) -2 ∉ N

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

a: \(=\dfrac{2a^2-6a+3a+9-3a^2-3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}\)

\(=\dfrac{-a^2-3a+6}{\left(a+3\right)}\cdot\dfrac{1}{â+1}=\dfrac{-a^2-3a+6}{\left(a+3\right)\left(a+1\right)}\)

b: |a|=2

=>a=2 hoặc a=-2

Khi a=2 thì \(A=\dfrac{-2^2-3\cdot2+6}{\left(2+3\right)\left(2+1\right)}=\dfrac{-4}{15}\)

Khi a=-2 thì \(A=\dfrac{-\left(-2\right)^2-3\cdot\left(-2\right)+6}{\left(-2+3\right)\left(-2+1\right)}=-8\)

a: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}-1}{3-\sqrt{a}}\)

\(=\dfrac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{2\sqrt{a}-9-a+9+2a-5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}-3\right)}\)

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

b: A là số nguyên

=>\(\sqrt{a}-3+2⋮\sqrt{a}-3\)

=>\(\sqrt{a}-3\in\left\{1;-1;2;-2\right\}\)

=>a thuộc {16;25;1}

\(\dfrac{5}{2}=\dfrac{1}{6}+\dfrac{b}{3}\)

\(\Leftrightarrow\dfrac{5}{2a}-\dfrac{1}{6}-\dfrac{b}{3}=0\)

msc : 18a

\(\Leftrightarrow\dfrac{45}{18a}-\dfrac{3a}{18a}-\dfrac{6ab}{18a}=0\)

\(\Leftrightarrow45-3a-6ab=0\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{15}{1+2b}\\b=\dfrac{15}{2a}-\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{5}{2a}\) hay \(\dfrac{5}{2}a\) vậy bạn?

\(\dfrac{5}{2a}=\dfrac{1}{6}+\dfrac{b}{3}\)

\(\Rightarrow\dfrac{15}{6a}=\dfrac{a}{6a}+\dfrac{2ab}{6a}\)

\(\Rightarrow a+2ab=15\)

\(\Rightarrow a\left(1+2b\right)=15\)

\(\Rightarrow a=\dfrac{15}{2b+1}\).

-Vì \(a,b\) nguyên nên để \(\dfrac{15}{2b+1}\) là số nguyên thì: \(15⋮\left(2b+1\right)\)

\(\Rightarrow2b+1\inƯ\left(15\right)\)

\(\Rightarrow2b+1\in\left\{1;3;5;15;-1;-3;-5;-15\right\}\)

\(\Rightarrow b\in\left\{0;1;2;7;-1;-2;-3;-8\right\}\)

*Với \(b=0\) thì \(a=\dfrac{15}{2.0+1}=15\)

*Với \(b=1\) thì \(a=\dfrac{15}{2.1+1}=5\)

*Với \(b=2\) thì \(a=\dfrac{15}{2.2+1}=3\)

*Với \(b=7\) thì \(a=\dfrac{15}{2.7+1}=1\)

*Với \(b=-1\) thì \(a=\dfrac{15}{2.\left(-1\right)+1}=-15\)

*Với \(b=-2\) thì \(a=\dfrac{15}{2.\left(-2\right)+1}=-5\)

*Với \(b=-3\) thì \(a=\dfrac{15}{2.\left(-3\right)+1}=-3\)

*Với \(b=-8\) thì \(a=\dfrac{15}{2.\left(-8\right)+1}=-1\)