Ta có: A = 3 + 3² + 3³ + ... + 3¹⁰⁰

=> 3A =  3² + 3³ + 3⁴ + ... + 3¹⁰¹

=> 3A - A = 3¹⁰¹ - 3

=> 2A = 3¹⁰¹ - 3

=> 2A + 3 = 3¹⁰¹

=> x = 101

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

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

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

\(2.3^{101}-6+3=3^x\)

\(3.\left(2.3^{100}-1\right)=3^x\)

\(=3^4-64-25\)

\(=-8\)

= 3⁴ - ( -8)² -  25

= 81 - 64 - 25

=  17 - 25

=    -8

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

\(\Rightarrow2x-1=3\)

     \(2x=3+1\)

  \(2x=4\)

 \(x=4:2=2\)

( 1/2 ) 2x -1= ( 1/2 ) 3 

2x - 1 =3 

2x = 3+1

2x=4

x= 4:2 = 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 ∈ {- \(\sqrt2\); \(\sqrt2\) )

b1:

\(-5.\left(x-\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\Leftrightarrow-5x+1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)

\(\Leftrightarrow-\frac{11}{2}x+\frac{4}{3}=\frac{3}{2}x-\frac{5}{6}\Leftrightarrow-\frac{11}{2}x-\frac{3}{2}x=-\frac{5}{6}-\frac{4}{3}\Leftrightarrow-7x=-\frac{13}{6}\)

\(\Leftrightarrow x=\frac{-13}{6}:\left(-7\right)=\frac{13}{42}\)

b2:

a)\(3^{3x-1}=9^{x-2}\Leftrightarrow3^{3x-1}=\left(3^2\right)^{x-2}\Leftrightarrow3^{3x-1}=3^{2x-4}\Leftrightarrow3x-1=2x-4\)

<=>3x-2x=-4+1<=>x=-3

b)\(\left(x-1\right)^4=\left(x-1\right)^6\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^6=0\Leftrightarrow\left(x-1\right)^4\left[1-\left(x-1\right)^2\right]=0\)

\(\Leftrightarrow\left(x-1\right)^4\left[1-\left(x^2-2x+1\right)\right]=0\Leftrightarrow\left(x-1\right)^4\left(1-x^2+2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^4\left(-x^2+2x\right)=0\Leftrightarrow\left(x-1\right)^4.\left(-x\right).\left(x-2\right)=0\)

<=>(x-1)⁴=0 hoặc -x=0 hoặc x-2=0 <=> x=1 hoặc x=0 hoặc x=2

\(3^x+25=26\cdot2^2+2\cdot3^0\)

\(3^x+25=26\cdot4+2\cdot1\)

\(3^x+25=104+2\)

\(3^x+25=106\)

\(3^x=106-25\)

\(3^x=81\)

\(3^x=3^4\)

Vậy : \(x=4\)

\(27\cdot3^x=243\)

\(3^x=\frac{243}{27}\)

\(3^x=9\)

\(3^x=3^2\)

Vậy x = 2

\(3^x+25=26\times2^2+2\times3^0\)

\(\Rightarrow3^x+25=26\times4+2\times1\)

\(\Rightarrow3^x+25=104+2\)

\(\Rightarrow3^x+25=106\)

\(\Rightarrow3^x=81\left(\text{cùng bớt đi 25}\right)\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

\(x^2-5x+4=0\)

\(x^2-x-4x+4=0\)

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

\(x.\left(x-1\right)-4.\left(x-1\right)=0\)

\(\left(x-4\right).\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)

x^2-5x+4=0

(x^2-x) - (4x-4) =0

x(x-1)-4(x-1)=0

(x-4)(x-1)=0

Ta có hai trường hợp:

x-4=0=) x=4

x-1=0=) x=1

( - 25 ) : 10 = -15 : x

=> -15 : x = - 25 : 10

     -15 : x = - 2,5

     x          = -15 : -2,5

     x          = 6

Vậy x = 6

Ta có: (-25) : 10 = (-15) : x

=>      -2,5          = (-15) : x

=>       x             = (-15) : (-2,5)

=>       x             = 6

Vậy x = 6