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

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

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

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

Vậy x = 1/2

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

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

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy x = 3 hoặc x = 1

\(\left(2x-1\right)^3=-8\)

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

\(\Leftrightarrow2x-1=-2\)

<=> 2x = -1

<=> x = -0,5

Vậy x = -0,5

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)

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

\(2x-1=-2\)

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

\(2x=-1\)

\(x=-1\times2\)

\(x=-2\)

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

\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

\(49\times7^x=2401\)

   \(7^x=2401:49\)

\(7^x=49\)

\(7^x=7^2\)

\(\Rightarrow x=2\)

\(2x^x-15=17\)

\(2x^x=17+15\)

\(2x^x=32\)

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

\(\Rightarrow x\times x=5\)

\(\Rightarrow x^2=5\)

\(2^x+3\times2^x=32\)

\(2^x\times\left(3+1\right)=32\)

\(2^x\times4=32\)

\(2^x=32:4\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

\(2x\left(x-\frac{1}{7}\right)=0\)

\(x\left(x-\frac{1}{7}\right)=0:2\)

\(x\left(x-\frac{1}{7}\right)=0\)

\(x\times x-x\times\frac{1}{7}=0\)

\(x^2-x\frac{1}{7}=0\)

\(x-x=0:\frac{1}{7}\)

\(x-x=0\)

\(\Rightarrow x=0\)

1. \(x^2+2y^2+2xy-2y+1=0\)

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

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

Có: \(\left(x+y\right)^2\ge0;\left(y-1\right)^2\ge0\)

Mà theo bài ra: \(\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

x¹³ = 27.x¹⁶

=> x¹³ - 27x¹⁶ = 0

=> x¹³(1 - 27x³) = 0

=> \(\orbr{\begin{cases}x^{13}=0\\1-27x^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\27x^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^3=\frac{1}{27}\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

c) \(\left(\frac{1}{2}\right)^{2x-1}=\frac{1}{8}\)

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

=> \(2x-1=3\)

=> \(2x=3+1\)

=> \(2x=4\)

=> \(x=4:2=2\)

\(b,\text{ }x^{13}=27\cdot x^{16}\)

\(x^{16}\text{ : }x^{13}=27\)

\(x^3=3^3\)

\(x=3\)

\(b,\text{ }x^{13}=27\cdot x^{16}\)

\(x^{16}\text{ : }x^{13}=27\)

\(x^3=3^3\)

\(x=3\)

có ai giải giúp mình ko mình đang gấp huhu

2^x x 16 = 128 

2^x          = 128 : 16

2 ^x        = 8

2^x         = 2³

3^x : 9 = 27 

3^x      = 27 x 9

3^x      =243

3^x      = 3⁵

[ 2^x + 1 ]³ = 27

2^x3 + 1³   = 27

2^x3 +1      = 27

2^x3           = 27 - 1

2^x3           = 26

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

\(\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)^3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{2};0\right\}\)

a)210-5x=200

5x=210-200

5x=10

x=10:2

x=5

B)210-5(x-10)=200

5(x-10)=210-200

5(x-10)=10

x-10=10:5=2

x=2+10=12

c)450:[41-(2x-5)]=3 mũ 2 .5

450:[41+5-2x]=45

41+5-2x=450:45

41+5-2x=10

2x=36

x=18

D)210:x-10=20

210:x=20+10

210:x=30

x=210:30

x=7

E)(5x-39).7+3=80

(5x-39).7=77

5x-39=11

5x=50

x=10

F(2x+1):7=2 mũ 2 nhân 3 mũ 2

(2x+1):7=36

2x+1=252

2x=251

x=251/2

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

\(\Rightarrow B=\frac{1}{2^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+...+\frac{1}{7.8}\)

\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{8^2}< 1-\frac{1}{2}+...+\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{8^2}< 1-\frac{1}{8}\)

\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{8^2}< \frac{7}{8}< 1\)

\(\Rightarrow B< 1\)

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