thanks bro ! :)

tích cho mk đi bro

Bài 1:

6) 3x + 2³ = 17 + 3²

3x + 8 = 17 + 9

3x + 8 = 26

3x = 26 - 8

3x = 18

x = 18 : 3

x = 6

Vậy x = 6

Bài 2:

3) 145 - (125 + x) = 12

125 + x = 145 - 12

125 + x = 133

x = 133 - 125

x = 8

Vậy x = 8

6) 3³ - (x - 5) = 2²

27 - (x - 5) = 4

x - 5 = 27 - 4

x - 5 = 23

x = 23 + 5

x = 28

Vậy x = 28

9) (x + 7) - 15⁰ = 202 - 19

(x + 7) - 1 = 189

x + 7 = 189 + 1

x + 7 = 190

x = 190 - 7

x - 183

Vậy x = 183

Ta có: \(\left(x-3\right)^3-3=3^0+3^1+2^5\cdot5\)

=>\(\left(x-3\right)^3-3=1+3+32\cdot5=160+4=164\)

=>\(\left(x-3\right)^3=167\)

=>\(x-3=\sqrt[3]{167}\)

=>\(x=3+\sqrt[3]{167}\)

\(\frac{25}{5^x}=5^x\)\(\Rightarrow x=1\)

Ta có: \(A=3+3^2+\cdots+3^{100}\)

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

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

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

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

=>\(3^{x+1}=3^{101}\)

=>x+1=101

=>x=100

a,   \(2^x-15=17\)

\(\Rightarrow2^x=17+15\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

b,   \(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=1000\)

\(\Rightarrow\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=10+11\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=21:7\)

\(\Rightarrow x=3\)

c,   \(x^{10}=1^x\)

\(\Rightarrow x\in\left\{1;0\right\}\)

\(2^x-15=17\)

\(\Rightarrow2^x=17+15\)

\(\Rightarrow2^x=32=2^4\)

\(\Rightarrow x=4\)

\(\left(7x-11\right)^3=2^5.5^2+200\)

Phần này mk ko bt làm đâu

\(x^{10}=1^x\)

\(\Rightarrow\)\(x^{10}=1\)

\(\Rightarrow x=1\)

\(x^{200}=x\)

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

\(x^{100}=1\)

\(\Rightarrow x=1\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow2x-15=2x-15\)

\(\Rightarrow x=1\)

\(\left(\frac{1}{3}\right)^{2x-1}-\frac{1}{3^2}=-\frac{2}{27}\)

=> \(\left(\frac{1}{3}\right)^{2x-1}=-\frac{2}{27}+\frac{1}{9}\)

=> \(\left(\frac{1}{3}\right)^{2x-1}=\frac{1}{27}\)

=> \(\left(\frac{1}{3}\right)^{2x-1}=\left(\frac{1}{3}\right)^3\)

=> 2x - 1 = 3

=> 2x = 3 + 1

=> 2x = 4

=> x = 4/2 = 2

\(\Leftrightarrow2^{x+1}.3^y=4^x.3^x\)

\(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)

\(\Leftrightarrow\frac{3^y}{3^x}=\frac{2^{2x}}{2^{x+1}}\)

\(\Leftrightarrow3^{y-x}=2^{x-1}\)

Nếu \(x>1\Rightarrow\) vế trái lẻ, vế phải chẵn pt vô nghiệm

\(\Rightarrow x=1\Rightarrow3^{y-1}=1\Rightarrow y=1\)

1: \(A=2+2^2+2^3+\cdots+2^{100}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{101}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{101}-2-2^2-2^3-\cdots-2^{100}\)

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

2: \(B=1+5+5^2+5^3+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

3: \(C=3+3^2+\cdots+3^{1000}\)

=>\(3C=3^2+3^3+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

Câu 1:

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

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

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

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

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

A = 0 + 0 + 0 + ... + 0 + 2\(^{101}\) - 2

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