a. (x - 2)² = 1

<=> (x - 2)² = 1² = (-1)²

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

Vậy x \(\in\){1; 3}.

b. (2x - 1)³ = -8

<=> (2x - 1)³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -2 + 1

<=> 2x = -1

<=> x = -1/2

Vậy x = -1/2.

c. (x + 1/2)² = 1/16

<=> (x + 1/2)² = (1/4)² = (-1/4)²

<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)

Vậy x \(\in\){-1/4; -3/4}.

d. (x - 2)³ = -27

<=> (x - 2)³ = (-3)³

<=> x - 2 = -3

<=> x = -3 + 2

<=> x = -1

Vậy x = -1.

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

<=> x-2=1 hoặc x-2=-1

<=> x= 3 hoặc x=1

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

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

2x-1=-2

2x=-1

x=-1/2

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

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

x+\(\frac{1}{2}\)=\(\frac{1}{4}\)  hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)

x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)

d.\(\left(x-2\right)^3\)=-27

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

x-2=-3

x=-1

g.\(3^{x-1}\)=\(\frac{1}{243}\)

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

\(3^{x-1}\)=\(3^{-5}\)

x-1=-5

x=-4

a) 3^x=27

3^x=3³

x=3

b.2^x:2^3=1

2^x:8=1

2^x=1x8

2^x=8

2^x=2³

x=3

c.27<3^x<243

3³=27(loại)

3⁴=81 (lấy)

3⁵=243(loại)

vậy x=4

d.4^x-^3=36

a) (X - 2)\(^2\) = 1 <=> X - 2 = \(\sqrt{1}\) <=> X = 1 + 2 <=> X = 3

Bài 1: 

a: \(\Leftrightarrow3^x\cdot10=810\)

\(\Leftrightarrow3^x=81\)

hay x=4

c: \(\Leftrightarrow5^x\cdot5+5^x\cdot\dfrac{1}{25}=126\)

\(\Leftrightarrow5^x\cdot\dfrac{126}{25}=126\)

\(\Leftrightarrow5^x=25\)

hay x=2

Bài 2: 

a: \(27^{11}=3^{33}\)

\(81^8=3^{32}\)

mà 33>32

nên \(27^{11}>81^8\)

c: \(625^5=\left(5^4\right)^5=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{21}\)

mà 20<21

nên \(625^5< 125^7\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

a, \(\left(2x-1\right)^2:9=49\)

\(\left(2x-1\right)^2=441\)

\(\Rightarrow\orbr{\begin{cases}2x-1=441\\2x-1=-441\end{cases}\Rightarrow\orbr{\begin{cases}x=221\\x=-220\end{cases}}}\)

b, \(3^x+3^{x+2}=810\)

\(3^x+3^x.3^2=810\)

\(3^x\left(1+3^2\right)=810\)

\(3^x.10=810\)

\(3^x=81=3^4\)

\(\Rightarrow x=4\)

\(\hept{\begin{cases}80⋮x\\56⋮x\end{cases}}\Rightarrow x\inƯC\left(80;56\right)\)

\(80=2^4.5\)

\(56=2^3.7\)

\(ƯCLN\left(80;56\right)=2^3=8\)

\(\RightarrowƯC\left(80;56\right)=Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Mà \(x\ge3\)

\(\Rightarrow x\in\left\{4;8\right\}\)

a) \(\left(2x-1\right)^2\div9=49\)

\(\Leftrightarrow\left(2x-1\right)^2=49\times9\)

\(\Leftrightarrow\left(2x-1\right)^2=441\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(\pm21\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=21\\2x-1=-21\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-10\end{cases}}\)

Vậy \(x\in\left\{-10;11\right\}\)

b) \(3^x+3^{x+2}=810\)

\(\Leftrightarrow3^x+3^x.3^2=810\)

\(\Leftrightarrow3^x\left(1+3^2\right)=810\)

\(\Leftrightarrow3^x.10=810\)

\(\Leftrightarrow3^x=810\div10\)

\(\Leftrightarrow3^x=81\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)