C/ Số số hạng của dãy trên là:

             (x - 1) + 1 = x (số hạng)

Tổng dãy trên là: x.(x + 1) / 2 = 55

=> x.(x + 1) = 55 x 2

=> x .(x + 1) = 110

=> x .(x + 1) = 10.11

=> x = 10

c) (x+1).x:2=55

(x+1).x=110

Tích của 2 số liên tiếp bằng 110

=>x=10

a)

\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)

b)

\(\frac{1}{4}-(2x-1)^2=0\)

\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)

\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)

c)

\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)

\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)

\(\Leftrightarrow 5-x=\frac{-3}{4}\)

\(\Leftrightarrow x=\frac{23}{4}\)

d)

\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)

\(\Rightarrow x=3,8:2=1,9\)

e)

\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)

\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)

f)

\(5^{(x+5)(x^2-4)}=1\)

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

\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)

g)

\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)

\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)

h)

\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)

\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)

a: \(\left(x-2,5\right)^2=-27\)

mà \(\left(x-2,5\right)^2\ge0\forall x\)

nên x∈∅

b: \(81^{x}:3^{x}=9\)

=>\(\left(\frac{81}{3}\right)^{x}=9\)

=>\(27^{x}=9\)

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

=>\(3^{3x}=3^2\)

=>3x=2

=>\(x=\frac23\)

c: \(\left(2x+1\right)^2=64\)

=>\(\left[\begin{array}{l}2x+1=8\\ 2x+1=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=7\\ 2x=-9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=-\frac92\end{array}\right.\)

d: \(2^{4-x}=16\)

=>\(2^{4-x}=2^4\)

=>4-x=4

=>x=4-4=0

bằng 1

a) \(\left(2x-1\right)^3=-8\)

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

=>  2x - 1 = -2

=>  x = -1/2

a,x=-1/2

b,x=0

c,x=2

d,x=4.5

e,x=2

f,x=1

g,x=...

k nhé

bài 12 :

a,\(\left(x-\frac{1}{2}\right)^2=0\)

Mà: 0²=0

=> \(\left(x-\frac{1}{2}\right)^2=0^2\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{2}\)

b,  \(\left(x-2\right)^2=1\)

Mà : 1=1²

\(\Rightarrow\left(x-2\right)^2=1^2\)

=> x - 2 = 1

=> x = 3

c, \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)=-2\)

Vì -8 =-2³

nên ...

=> 2x =-1

=> x=0.5

d.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

cái này cũng như mấy cái trên thôi

Bài 12:

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(x-2=\pm1\)

• Nếu \(x-2=1\)

\(x=3\)

• Nếu \(x-2=-1\)

\(x=1\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow2x-1=-2\)

\(2x=-1\)

\(x=-\frac{1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(x+\frac{1}{12}=\pm\frac{1}{4}\)

• Nếu \(x+\frac{1}{12}=\frac{1}{4}\)

\(x=\frac{1}{6}\)

• Nếu \(x+\frac{1}{12}=-\frac{1}{4}\)

\(x=-\frac{1}{3}\)

Bài 13: có người làm rồi

Bài 14:

a) \(25^3\div5^2\)

\(=\left(5^2\right)^3\div5^2\)

\(=5^6\div5^2=5^4\)

b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)

\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)

\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)

c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)

\(=3-1+\frac{1}{4}:2\)

\(=2+\frac{1}{8}=2\frac{1}{8}\)

(x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

(2x - 1)³ = -64

=> (2x - 1)³ = -4³

=> 2x - 1 = -4

=> 2x = -4 + 1

=> 2x = -3

=> x = -3/2

( x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

2.(2\(x\) + 2)\(^2\) = 32

(2\(x\) + 2)\(^2\) = 32 : 2

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

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

2\(x\) + 2 = 4

2\(x\) = 4 - 2

2\(x\) = 2

\(x\) = 2 : 2

\(x=1\)

Vậy \(x\) = 1

\(2\cdot\left(2x+2\right)^2=32\)

=>\(\left(2x+2\right)^2=\frac{32}{2}=16\)

=>\(\left[\begin{array}{l}2x+2=4\\ 2x+2=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=2\\ 2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\end{array}\right.\)

ta có

\(A \left(\right. x \left.\right) = - 2 x^{2} - 3 x^{6} - 0.01 = 0\)

\(- 2 x^{2} - 3 x^{6} = 0.01\)

\(- 2 x^{2} - 3 x^{6} \leq 0\) (vì \(- 2 x^{2} \leq 0\) và \(- 3 x^{6} \leq 0\))

Vế phải 0.01 > 0\(\)

Một số ko âm không thể bằng một số dương

Vậy phương trình vô nghiệm

Ta có: \(3x^6\ge0\forall x\)

\(2x^2\ge0\forall x\)

Do đó: \(3x^6+2x^2\ge0\forall x\)

=>\(-3x^6-2x^2\le0\forall x\)

=>\(A=-3x^6-2x^2-0,01\le-0,01<0\forall x\)

=>A không có nghiệm