8n-7 chia hết cho 2n+1

=>4(2n+1)-11 chia hết cho 2n+1

=>2n+1 thuộc Ư(11)={1;11}

+/2n+1=1=>2n=0=>n=0

+/2n+1=11=>2n=10=>n=5

=> n thuộc {0;5}

Yay ~ Sáng mai tốt quá :v 9h là rảnh rồi

a) Sửa \(Ba\left(PO_4\right)_2\rightarrow Ba_3\left(PO_4\right)_2\)

\(3Ba\left(OH\right)_2+2H_3PO_4\rightarrow Ba_3\left(PO_4\right)_2+6H_2O\)

b) \(3Cu\left(OH\right)_2+2H_3PO_4\rightarrow Cu_3\left(PO_4\right)_2+6H_2O\)

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

Pt: \(A_2O_a+aH_2\rightarrow2A+aH_2O\)

\(2M_A+16a\) a

13,38g 0,06mol

\(\Rightarrow\dfrac{2M_A+16a}{13,38}=\dfrac{a}{0,06}\)

\(\Rightarrow M_A=103,5a\)

loại nhận loại

A là Pb

a) \(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)\)

\(=\left[\left(2x^3+10x\right)+\left(x^4-25\right)\right]:\left(x^2+5\right)\)

\(=\left[2x\left(x^2+5\right)+\left(x^2-5\right)\left(x^2+5\right)\right]:\left(x^2+5\right)\)

\(=\left(x^2+5\right)\left(x^2+2x-5\right):\left(x^2+5\right)\)

\(=x^2+2x-5\)

2034=2.3².113

a) \(2\left(x+5\right)-4x=1\)

\(\Leftrightarrow-2x+9=0\)

\(\Rightarrow x=\dfrac{9}{2}\)

b) \(\left(x-4\right)\left(5x-2\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(5x-2\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(5x-2+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\5x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-1}{5}\end{matrix}\right.\)

Vậy..

c) \(x^4-8x^2-9=0\)

\(\Leftrightarrow\left(x^2\right)^2-2x^24+16-25=0\)

\(\Leftrightarrow\left(x^2-4\right)^2-5^2=0\)

\(\Leftrightarrow\left(x^2-4-5\right)\left(x^2-4+5\right)=0\)

\(\Leftrightarrow\left(x^2-9\right)\left(x^2+1\right)=0\)

Vì x² + 1 > 0

\(\Rightarrow x^2-9=0\)

\(\Rightarrow x=\pm3\)