chuẩn đoán mạng 

a) 4,324m³ = 4324dm³     19,25m³ = 19250dm³     8dm³ 512cm³ = 8512cm³
12500cm³ = 12,5dm³       
b) 9,512dm³ = 9512cm³     0,8dm³ = 800cm³            7,09dm³ = 7090cm³          
\(\dfrac{1}{4}\)m³ = 250000cm³

Bài 2: 

a: 2/3+1/2-1/6=4/6+3/6-1/6=3/6+3/6=1

b: \(=\dfrac{12}{24}\cdot\dfrac{23}{46}\cdot\dfrac{1}{2}=\dfrac{1}{4}\cdot\dfrac{1}{2}=\dfrac{1}{8}\)

c: =1/6x1/3=1/18

d: \(=\dfrac{48}{24}\cdot23=46\)

\(\dfrac{1}{8}-\dfrac{4}{8}=\dfrac{-3}{8}\)

-3/8

1/8 -4/8 = 1/8 + ( - 4/8 ) =-3/8

Bài 1:

\(A=3+3^2+...+3^{100}\)

=>\(3\cdot A=3^2+3^3+...+3^{101}\)

=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)

=>\(2A=3^{101}-3\)

=>\(2A+3=3^{101}\)

mà \(2A+3=3^n\)

nên n=101

 Bài 2:

a: \(M=3+3^2+3^3+3^4+...+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{98}\right)⋮12\)

=>\(M=4\cdot3\cdot\left(1+3^2+...+3^{98}\right)⋮4\)

b: \(M=3+3^2+...+3^{100}\)

=>\(3M=3^2+3^3+...+3^{101}\)

=>\(3M-M=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)

=>\(2M=3^{101}-3\)

=>\(2M+3=3^{101}\)

=>n=101

a: Xét (O) có 

ΔABC nội tiếp

AB là đường kính

Do đó: ΔABC vuông tại C

It is common knowledge that you have to drink more fluids when you have flu

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