a) Ta có: \(M_{XCO_3}=4\cdot25=100\) \(\Rightarrow M_X=100-12-16\cdot3=40\left(đvC\right)\)

\(\Rightarrow\) X là Canxi (Ca)

b) \(\%O=\dfrac{16\cdot3}{100}\cdot100\%=48\%\)

Bài 3:

a) M(XCO3)=25. M(He)= 25.4=100(đ.v.C)

Mặt khác: M(XCO3)=M(X)+ 60

=> M(X)+60=100

<=>M(X)=40(đ.v.C) 

=> X là Canxi (Ca=40)

b) %mO=[(3.16)/100].100=48%

Chọn D

1+2+2²+..........+2²⁰⁰⁹+2²⁰¹⁰

=(1+2+2²)+.........+(2²⁰⁰⁷+2²⁰⁰⁸+2²⁰⁰⁹)+2²⁰¹⁰

=7+..........+2²⁰⁰⁷.(1+2+2²)+2²⁰¹⁰

=7+..........+2²⁰⁰⁷.7+2²⁰¹⁰

=>A chia 7 dư 2²⁰¹⁰

Ta có:2³=8 đồng dư với 1(mod 7)

=>(2³)⁶⁷⁰=2²⁰¹⁰ đồng dư với 1⁶⁷⁰(mod 7)

=>2²⁰¹⁰ đồng dư với 1(mod 7)

=>2²⁰¹⁰ chia 7 dư 1

=>A chia 7 dư 1

giải chi tiết ra giúp mk !

\(5x\left(x-2018\right)-x+2018=0\)

\(5x\left(x-2018\right)-\left(x-2018\right)=0\)

\(\left(x-2018\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2018=0\\5x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{5}\end{cases}}\)

Vậy.........

x = 2018

x = 1/5

t i c k nha

\(5x\left(x-2018\right)-x+2018=0\)

\(\Leftrightarrow5x\left(x-2018\right)-\left(x-2018\right)=0\)

\(\Leftrightarrow\left(x-2018\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2018=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{5}\end{cases}}\)

Mà h cóa ai bik làm ko giúp mk dới :)) 

\(D=\dfrac{11}{4}\cdot\left(-0,4\right)-\dfrac{8}{5}\cdot2,75+\left(-1,2\right):\dfrac{4}{11}\)

\(D=\dfrac{11}{4}\cdot\left(\dfrac{-2}{5}\right)-\dfrac{8}{5}\cdot\dfrac{11}{4}+\left(\dfrac{-6}{5}\right)\cdot\dfrac{11}{4}\)

\(D=\dfrac{11}{4}\cdot\) \(\left(\dfrac{-2}{5}-\dfrac{8}{5}+\dfrac{-6}{5}\right)\)

\(D=\dfrac{11}{4}\cdot\left(-\dfrac{16}{5}\right)\)

\(D=-\dfrac{44}{5}\)

a)

$n_{Nito} = \dfrac{6,02.10^{23}}{6,02.10^{23}} = 1(mol)$
$m_{Nito} = 1.14 = 14(gam)$

b)

$n_{Cl} = \dfrac{6,02.10^{23}}{6,02.10^{23}} = 1(mol)$

$m_{Cl} = 1.35,5 = 35,5(gam)$

c)

$n_{H_2O} = \dfrac{6,02.10^{23}}{6,02.10^{23}} = 1(mol)$

$m_{H_2O} = 1.18 = 18(gam)$

a) \(n_{N_2}=\dfrac{6,02.10^{23}}{6,02.10^{23}}=1\left(mol\right)\)

=> \(m_{N_2}=1.28=28\left(g\right)\)

b)  \(n_{Cl_2}=\dfrac{6,02.10^{23}}{6,02.10^{23}}=1\left(mol\right)\)

=> \(m_{Cl_2}=1.35,5.2=71\left(g\right)\)

c)  \(n_{H_2O}=\dfrac{6,02.10^{23}}{6,02.10^{23}}=1\left(mol\right)\)

=> \(m_{H_2O}=1.18=18\left(g\right)\)

d)  \(n_{CaCO_3}=\dfrac{6,02.10^{23}}{6,02.10^{23}}=1\left(mol\right)\)

=> \(m_{CaCO_3}=1.100=100\left(g\right)\)

Kẻ Oz//Ax//By

=>góc zOA=góc xAO=35 độ

Oz//By

=>góc zOB+góc yBO=180 độ

=>góc zOB=50 độ

=>góc AOB=35+50=85 độ

Bn chỉ cần tách ra thôi là lm dc