GIÚP MÌNH VỚI Ạ MÌNH CẦN GẤP

1: x∈B(3)

=>x∈{3;6;9;12;15;18;21;...}

mà 12<=x<=18

nên x∈{12;15;18}

2: 2+4+6+...+2x=210

=>1+2+...+x=105

=>\(x\cdot\frac{\left(x+1\right)}{2}=105\)

=>x(x+1)=210

=>\(x^2+x-210=0\)

=>(x+15)(x-14)=0

=>x=-15(loại) hoặc x=14(nhận)

3: \(\left(x-2\right)^6=\left(x-2\right)^8\)

=>\(\left(x-2\right)^8-\left(x-2\right)^6=0\)

=>\(\left(x-2\right)^6\cdot\left\lbrack\left(x-2\right)^2-1\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(x-2\right)^6=0\\ \left(x-2\right)^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x-2=0\\ x-2=1\\ x-2=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\\ x=1\end{array}\right.\)

4: 96-3(x+1)=42

=>3(x+1)=96-42=54

=>x+1=18

=>x=17

5: 15x-9x+2x=72

=>\(x\cdot\left(15-9+2\right)=72\)

=>8x=72

=>x=72/8

=>x=9

6: \(3^{x+2}+3^{x}=10\)

=>\(3^{x}\left(3^2+1\right)=10\)

=>\(3^{x}\cdot10=10\)

=>\(3^{x}=1\)

=>x=0

1, Tính :

a, \(8.4^5-3.9.12+28\div4.2^0\)

\(=8192-27.12+7.1\)

\(=8192-324+7\)

\(=7868+7\)

\(=7875\)

b, \(210+86.2^3\div2-19.2^3\)

\(=210+86.8\div2-19.8\)

\(=210+688\div2-19.8\)

\(=210+344-152\)

\(=554-152=402\)

c, \(54.23+11^2-152\)

\(=54.23+121-152\)

\(=1242+121-152\)

\(=1363-152=1211\)

2, Tính giá trị của biểu thức :

a, \(311\div\left\{210+\left[53+\left(37-25\right).2^2\right]\right\}\)

\(=311\div\left\{210+\left[53+12.4\right]\right\}\)

\(=311\div\left\{210+\left[53+48\right]\right\}\)

\(=311\div\left\{210+101\right\}\)

\(=311\div311=1\)

Vậy giá trị biểu thức trên bằng 1

b, \(\left\{\left[200+\left(50-30\right)^2\right]-456\right\}\div12\)

\(=\left\{\left[200+20^2\right]-456\right\}\div12\)

\(=\left\{\left[200+400\right]-456\right\}\div12\)

\(=\left\{600-456\right\}\div12\)

\(=144\div12=12\)

Vậy giá trị biểu thức trên bằng 12

1. Ta có: A = 3⁰ + 3¹ + 3² + ... + 3¹⁰⁰

3A = 3.(1 + 3 + 3² + ... + 3¹⁰⁰)

3A = 3 + 3² + 3³ + ... + 3¹⁰¹

3A - A = (3 + 3² + 3³ + ... + 3¹⁰¹) - (1 + 3 + 3² + ... + 3¹⁰⁰)

2A = 3¹⁰¹ - 1

A = \(\frac{3^{101}-1}{2}\)

Vậy ...

Baif1 :

đặt \(A=3^0+3^1+3^2+...+3^{100}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{101}\)

\(\Rightarrow3A-A=\left(3+3^2+...+3^{101}\right)-\left(1+3+...+3^{100}\right)\)

\(\Rightarrow2A=3^{101}-1\)

\(\Rightarrow A=\frac{3^{101}-1}{2}\)

\(\)3²ⁿ và 2³ⁿ

ta có 3ⁿ²=(3²)ⁿ⁼9ⁿ

2³ⁿ=(2³)ⁿ = 8ⁿ

mà 9>8

=>9ⁿ>8ⁿ hay 3²ⁿ>2³ⁿ

3²¹⁰ và 2³⁵⁰

Ta có: 3²¹⁰ = 3^3.70 = (3³)⁷⁰ = 27⁷⁰

2³⁵⁰ = 2^5.70 = (2⁵)⁷⁰ = 32⁷⁰

Mà 27 < 32

\(\Rightarrow\) 27⁷⁰ < 32⁷⁰

hay 3²¹⁰ < 2³⁵⁰