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

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

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

Teachers give more freedom to their students in classroom activities now

a) \(A=\left(x+5\right)^2-\left(x+3\right)^2\)

\(=\left[\left(x+5\right)-\left(x+3\right)\right]\left[\left(x+5\right)+\left(x+3\right)\right]\)

\(=\left(x+5-x-3\right)\left(x+5+x+3\right)\)

\(=2\left(2x+8\right)\)

\(=4x+16\)

b) \(B=\left(4x+1\right)^2-\left(2x+1\right)^2\)

\(=\left[\left(4x+1\right)-\left(2x+1\right)\right]\left[\left(4x+1\right)+\left(2x+1\right)\right]\)

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

\(=2x\left(6x+2\right)\)

\(=12x^2+4x\)

c) \(C=\left(3-4x\right)^2-\left(2x-1\right)\left(8x-9\right)\)

\(=9-24x+16x^2-16x^2+18x+8x-9\)

\(=\left(16x^2-16x^2\right)+\left(-24x+18x+8x\right)+\left(9-9\right)\)

\(=2x\)

d) \(D=\left(4+2x^2\right)-\left(1-4x\right)\left(4-x\right)\)

\(=4+2x^2-4+x+16x-4x^2\)

\(=\left(2x^2-4x^2\right)+\left(x+16x\right)+\left(4-4\right)\)

\(=-2x^2+17x\)

e) \(E=\left(2-3x\right)^2-2\left(2-3x\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left(2-3x+3x+5\right)^2\)

\(=7^2\)

\(=49\)

Bài 1:

a: \(k=\dfrac{y}{x}=\dfrac{y_1}{x_1}=\dfrac{2}{3}\)

=>\(y=\dfrac{2}{3}x\)

b: Khi \(x_2=9\) thì \(y_2=\dfrac{2}{3}\cdot x_2=\dfrac{2}{3}\cdot9=6\)

c: Khi \(y_3=8\) thì \(\dfrac{2}{3}\cdot x_3=8\)

=>\(x_3=8:\dfrac{2}{3}=12\)

Bài 2:

a: x,y tỉ lệ thuận

nên \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)

=>\(\dfrac{6}{y_2}=\dfrac{4}{9}\)

=>\(y_2=6\cdot\dfrac{9}{4}=\dfrac{3}{2}\cdot9=\dfrac{27}{2}\)

b: \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)

=>\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\dfrac{x_1+x_2}{y_1+y_2}=\dfrac{2}{-10}=-\dfrac{1}{5}\)

=>\(\dfrac{x}{y}=-\dfrac{1}{5}\)

=>y=-5x

grsgrs

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