\(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+3x+5\right)=0\)

mà \(x^2+3x+5=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\forall x\)

nên x-2=0

=>x=2

\(4x^4+81\)

\(=4x^4+36x^2+81-36x^2\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)

=>a=2; b=6

a+b=2+6=8

\(-5x^3+xy^2z^3\) có bậc là 1 + 2 + 3 = 6 

\(-5x^3+xy^2z^3\) có bậc là \(MAX\left(3;1+2+3\right)=1+2+3=6\)

a) Ta có: \(a^{2k}=5\)

\(P=2a^{6k}-4=2\cdot a^{3\cdot2k}-4=2\cdot\left(a^{2k}\right)^3-4\)

\(=2\cdot5^3-4=2\cdot125-4=250-4=246\) 

b) Ta có: \(a^{3k}=-5\)

\(Q=2a^{6k}-4=2\cdot a^{3k\cdot2}-4=2\cdot\left(a^{3k}\right)^2-4\\ =2\cdot\left(-5\right)^2-4=2\cdot25-4=50-4=46\)

\(\left(2x^2z^2\right)^3+\left(-3xy^3\right)^2=0\)

=>\(8x^6z^6+9x^2y^6=0\)

=>\(x^2\left(8x^4z^6+9y^6\right)=0\)

=>\(\left\{{}\begin{matrix}x^2=0\\8x^4z^6+9y^6=0\end{matrix}\right.\)

=>x=y=0

1 she prefers reading books to watching movies
2 why dont we try a new recipe for dinner?
3 it takes us two days to go camping in the countryside
4 he is too short to reach the top shelf without needing a ladder

a) \(Al_2O_3+H_2SO_4\rightarrow X+H_2O\)

X là chất Al₂(SO₄)₃

b) \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

c) áp dụng công thức định luật bảo toàn khối lượng

\(m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2O}-m_{Al_2O_3}=34,2+5,4-10,1=29,5\left(g\right)\)

vậy a = 29,5 g

9:

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

\(P=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

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

b: Khi x=2 thì \(P=\dfrac{2-1}{2}=\dfrac{1}{2}\)

c: \(S=P\cdot\dfrac{2}{x-2}=\dfrac{x-1}{2}\cdot\dfrac{2}{x-2}=\dfrac{x-1}{x-2}=\dfrac{x-2+1}{x-2}=1+\dfrac{1}{x-2}\)

Để S là số nguyên thì \(x-2\in\left\{1;-1\right\}\)

=>\(x\in\left\{3;1\right\}\)

mình cảm ơn ạ