Bài 1 :

\(2^x.8=512\)

\(2^x=512:8\)

\(2^x=64\)

\(2^x=2^6\)

\(\Rightarrow x=6\)

\(b,\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

\(c,x^{20}=x\)

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

\(d,\left(x-3\right)^{10}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

1. Tính:

B= 12 . 108+ 17 . 107

B = 1296 + 1819

B = 3115

2. Tìm x:

a) 4x = 2x+1

4x - 2x = 1

2x = 1

x = 1/2

Vậy x = 1/2

b) 16 = (x-1)^4

2^4 = (x - 1)^4

x - 1 = - 2 hoặc x - 1 = 2

x - 1 = -2, x = -2 + 1, x = -1

x - 1 = 2, x = 2 + 1, x = 3

Vậy x ∈ {-1; 3}

Tìm x,biết :

câu 1: (x+8)^2=7225

(x + 8)^2 = 85^2

x + 8 = -85 hoặc x + 8 = 85

x + 8 = - 85

x = - 85 - 8

x = - 93

x + 8 = 85

x = 85 - 8

x = 77

Vậy x ∈ {-93; 77}

Câu 2: x^2-190=710

x^2 = 710 + 190

x^2 = 900

x^2 = 30^2

x = -30 hoặc x = 30

Vậy x ∈ {-30; 30}

a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)

\(\left(2x+1\right)^2=\frac{25}{16}\)

\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)

\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)

\(x^3-\frac{9}{16}.x=0\)

\(x\left(x^2-\frac{9}{16}\right)=0\)

\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)

Bài 1:

a, x = 0

b, x = 2

c, x = 1

Bài 2:

Nếu n=0 thì nên bới giá trị -n²

bạn giải ra giúp mình với