Mk giúp pn bài 1 thui nha...

a)  A=3+3²+3³+...+3¹⁰⁰

<=>A=(3+3²) +(3³+3⁴) +...+(3⁹⁹+3¹⁰⁰)

<=>A=12+3².(3+3²)+...+3⁹⁸.(3+3²)

<=>A=12+3².12+...+3⁹⁸.12

<=>A=12.(3²+3³+...+3⁹⁸)

Ta có 12 chia hết cho 4 => 12.(3²+3³+...+3⁹⁸) chia hết cho 4 => A chia hết cho 4

Vậy A chia hết cho 4

b) A=3+3²+3³+...+3¹⁰⁰

<=> 3A=3²+3³+...+3¹⁰¹

<=>3A-A=3²+3³+...+3¹⁰¹-3-3²-3³-...-3¹⁰⁰

<=>2A=3¹⁰¹-3

<=>A=(3¹⁰¹-3)/2

Thay A=(3¹⁰¹-3)/2 vào 2A+3=3^x-1 ta có:

2.[(3¹⁰¹-3)/2]+3=3^x-1

<=>3¹⁰¹-3+3=3^x-1

<=>3¹⁰¹=3^x-1

<=>x-1=101

<=>x=102

vậy x=102

Ai thấy đúng tích nha , mấy pn kb +theo dõi mk vs ạ....

a) \(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

b) \(2^x.16=128\)

\(2^x=128:16\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

c) \(3^x:9=27\)

\(3^x=27.9\)

\(3^x=243\)

\(3^x=3^5\)

\(\Rightarrow x=5\)

d) \(x^4=x\)

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

e) \(\left(2x+1\right)^3=27\)

\(\left(2x+1\right)^3=3^3\)

\(\Rightarrow2x+1=3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

f) \(\left(x-2\right)^2=\left(x-2\right)^4\)

\(\left(x-2\right)^2-\left(x-2\right)^4=0\)

\(\left(x-2\right)^2-\left(x-2\right)^2.\left(x-2\right)^2=0\)

\(\left(x-2\right)^2\left[1-\left(x-2\right)^2\right]=0\)

\(\left(x-2\right)^2\left(1-x+2\right)\left(1+x-2\right)=0\)

\(\Rightarrow\left(x-2\right)^2=0\)hoac \(\orbr{\begin{cases}3-x=0\\x-1=0\end{cases}}\)

\(\Rightarrow x-2=0\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

a) \(3^x=81\Leftrightarrow3^x=3^4\Rightarrow x=4\)

b)\(2^x\times16=128\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)

c) \(3^x\div9=27\Leftrightarrow3^x\div3^2=3^3\Rightarrow x=5\)

d) \(x^4=x\Leftrightarrow x=1\)

e) \(\left(2x+1\right)^3=27\Leftrightarrow\left(2x+1\right)^3=3^3\Rightarrow2x+1=3 \)

\(\Rightarrow2x=3+1\Leftrightarrow2x=4\Rightarrow x=2\)

F) 

a) \(\frac{x-2}{3}=\frac{x+1}{4}\)

=> (x - 2).4 = 3.(x + 1)

=> 4x - 8 = 3x + 3

=> 4x - 3x = 3 + 8

=> x = 11

Vậy x = 11

b) \(2.\left(x+3\right)-\frac{1}{2}=x-1\)

=> \(2x+6-\frac{1}{2}=x-1\)

=> \(2x+\frac{11}{2}=x-1\)

=> \(2x-x=-1-\frac{11}{2}\)

=> \(x=-\frac{13}{2}\)

Vậy \(x=-\frac{13}{2}\)

Tìm \(x\) câu a:

\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0

\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0

\(\frac{11}{15}x\) = - \(\frac25\)

\(x=-\frac25:\frac{11}{15}\)

\(x\) = - \(\frac25\times\frac{15}{11}\)

\(x\) = - \(\frac{6}{11}\)

Vậy \(x=-\frac{6}{11}\)

Tìm \(x\) câu b:

\(x\) x 25% = 0,5

\(x\times0,25\) = 0,5

\(x=0,5:0,25\)

\(x=2\)

Vậy \(x=2\)

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

1/3x + 2/5x = - 2/5

x(1/3 + 2/5) = -2/5

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

1/3x + 2/5x = - 2/5

x(1/3 + 2/5) = -2/5

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )