(x-1)(x-2)>0

=>x-1 và x-2 cùng dấu

TH1: \(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>2\end{cases}\Rightarrow}x>2}\)

TH2: \(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 1\\x< 2\end{cases}\Rightarrow}x< 1}\)

bạn hc chương trình mới hay trương trình cũ

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

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

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

2=3

<=>2+10-8=3+15-12

<=>2.(1+5-4)=3.(1+5-4)

<=>2=3 (cùng bớt đi 1+5-4)

=>đpcm

ko chắc nữa

ơ

2+10-8 có bằng 3+15-12 đâu

ai tick mình mình tick cho

Bạn phóng to anh lên để mik giải

\(2R+H_2SO_4\rightarrow R_2SO_4+H_2\left(1\right)\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{H_2SO_4\left(2\right)}=\dfrac{1}{2}n_{NaOH}=0,015\left(mol\right)\\ \Rightarrow n_{H_2SO_4\left(1\right)}=0,075-0,015=0,06\left(mol\right)\\ n_R=2n_{H_2SO_4\left(1\right)}=0,12\left(mol\right)\\ \Rightarrow M_R=\dfrac{4,68}{0,12}=39\left(Kali-K\right)\)

Ta có: \(n_{H_2SO_{4_{tham.gia}}}=\dfrac{250}{1000}.0,3=0,075\left(mol\right)\)

\(n_{NaOH}=\dfrac{60}{1000}.0,5=0,03\left(mol\right)\)

\(PTHH:\)

\(2R+H_2SO_4--->R_2SO_4+H_2\left(1\right)\)

\(H_2SO_4+2NaOH--->Na_2SO_4+2H_2O\left(2\right)\)

Theo PT₍₂₎: \(n_{H_2SO_{4_{dư}}}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,03=0,015\left(mol\right)\)

\(\Rightarrow n_{H_2SO_{4_{PỨ}}}=0,075-0,015=0,06\left(mol\right)\)

Theo PT₍₁₎: \(n_R=2.n_{H_2SO_4}=2.0,06=0,12\left(mol\right)\)

\(\Rightarrow M_R=\dfrac{4,68}{0,12}=39\left(\dfrac{g}{mol}\right)\)

Vậy R là kim loại kali (K)