a: \(1=4^0\)

\(4=4^1\)

\(16=4^2\)

\(256=4^4\)

b: \(\dfrac{1}{4}=4^{-1}\)

\(\dfrac{1}{64}=4^{-3}\)

\(\dfrac{1}{256}=4^{-4}\)

\(\dfrac{1}{16}=4^{-2}\)

\(\dfrac{1}{1024}=4^{-5}\)

2n-1 là Ư4   mà Ư4 = 1,-1,2,-2,4,-4

-> n= 1,0

4n+ 1 là Ư10 mà Ư10= -1,1,2,-2,5,-5,10,-10

-> n= 0,1

(bài này mk làm nếu n là số nguyên nhé)

cảm ơn bạn nha

b 2n- là ước của 4 

c 4n+1 là ước của 10 

d 3n+1 chia n -2 

e 5n chia n + 1 

f n+8 chia n +1 

B4:

\(CTTQ:Na_xS_yO_z\left(x,y,z:nguy\text{ê}n,d\text{ươ}ng\right)\\ n_{Na}=\dfrac{4.6}{23}=0,2\left(mol\right);n_S=\dfrac{3,2}{32}=0,1\left(mol\right);n_O=\dfrac{4,8}{16}=0,3\left(mol\right)\\ x:y:z=0,2:0,1:0,3=2:1:3\\ \Rightarrow x=2;y=1;z=3\\ \Rightarrow CTHH:Na_2SO_3\)

Mình cần phải có tư duy, chớ nên phụ thuộc vào người khác quá nhiều nhé!

Bài 3:

\(CTTQ:Al_xS_yO_z\left(x,y,z:nguy\text{ên},d\text{ươ}ng\right)\\ x:y:z=\dfrac{15,8\%}{27}:\dfrac{28,1\%}{32}:\dfrac{56,1\%}{16}\approx0,0059:0,0087:0,0351\approx2:3:12\\ \Rightarrow CT\text{Đ}GN:Al_2S_3O_{12}\)

Nó cứ sai sai

Bài 2:

\(\text{Đ}\text{ặt}:N_xH_y\left(x,y:nguy\text{ê}n,d\text{ươ}ng\right)\\ M_{hc}=M_{H_2}.8,5=2.8,5=17\left(\dfrac{g}{mol}\right)\\ x=\dfrac{82.35\%.17}{14}=1\left(mol\right);y=\dfrac{17,65\%.17}{1}=3\left(mol\right)\\ CTHH:NH_3\\ b,n_{nguy\text{ê}n.t\text{ử}}=0,5.1+0,5.3=2\left(mol\right)\)

ta có: \(A=\frac{4n+1}{2n+3}=\frac{4n+6-5}{2n+3}=\frac{2.\left(2n+3\right)-5}{2n+3}=2-\frac{5}{2n+3}\)

Để A thuộc Z

=> 5/2n+3 thuộc Z

=> 5 chia hết cho 2n +3

=> 2n+3 thuộc Ư(5)={1;-1;5;-5}

nếu 2n + 3 = 1 => 2n = -2 => n = -1 (Loại)

2n+3 = -1 => 2n=-4 => n = -2 (Loại)

2n+3 = 5 => 2n = 2 => n = 1 (TM)

2n+3 = -5 => 2n = -8 => n = -4 (Loại)

\(\Rightarrow n\ne1\) thì A là phân số ( n thuộc N)

Cảm ơn bạn CÔNG CHÚA ÔRI nhiều ạ

(4n+7)⋮(4n+1)

Vì 4n+7=(4n+1)+6

=>(4n+1)+6⋮(4n+1) mà 4n+1⋮4n+1

=>6⋮4n+1

=>4n+1∈Ư(6)={1;2;3;6}

=>4n∈{0;1;2;5}

=>n∈{0;0,25;0,5;1,25}

Vì n là stn nên n=0

\(B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{1}{5}B=\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)

\(B-\frac{1}{5}B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}-\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\)\(\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{4}{5}B=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{4}{5}B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+\frac{4}{11}-\frac{4}{15}+...+\frac{4}{4n-1}-\frac{4}{4n+3}\)

\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{4n-3}\)

\(\frac{4}{5}B=\frac{16n-24}{12n-9}\)

\(B=\frac{\frac{16n-24}{12n-9}}{\frac{4}{5}}\)

\(B=\frac{20n-30}{12n-9}\)

B = \(\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}+\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}.\left(\frac{1}{3}-\frac{1}{4n+3}\right)=\frac{5}{12}-\frac{5}{4\left(4n+3\right)}=\frac{5}{12}-\frac{5}{16n+12}\)

sửa lại

\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)

\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{4n+3}\)

\(\frac{4}{5}B=\frac{4n}{12n+9}\)

\(B=\frac{\frac{4n}{12n+9}}{\frac{4}{5}}\)

\(B=\frac{5n}{12n+9}\)