`@` `\text {Ans}`

`\downarrow`

\(\left(2\cdot x+2\right)^2=64\)

`\Rightarrow`\(\left(2x+2\right)^2=\left(\pm8\right)^2\)

`\Rightarrow`\(\left[{}\begin{matrix}2x+2=8\\2x+2=-8\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=8+2\\2x=-8+2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=10\\2x=-6\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=10\div2\\x=-6\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

`@` `\text {Kaizuu lv uuu}`

\(2^x+3.4^2=64\\ \Rightarrow2^x+48=64\\ \Rightarrow2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\)

\(\Rightarrow2^x+16=64\\ \Rightarrow2^x=48\Rightarrow x\in\varnothing\)

a: x^3=7^3

=>x^3=343

=>\(x=\sqrt[3]{343}=7\)

b: x^3=27

=>x^3=3^3

=>x=3

c: x^3=125

=>x^3=5^3

=>x=5

d: (x+1)^3=125

=>x+1=5

=>x=4

e: (x-2)^3=2^3

=>x-2=2

=>x=4

f: (x-2)^3=8

=>x-2=2

=>x=4

h: (x+2)^2=64

=>x+2=8 hoặc x+2=-8

=>x=6 hoặc x=-10

j: =>x-3=2 hoặc x-3=-2

=>x=1 hoặc x=5

k:

9x^2=36

=>x^2=36/9

=>x^2=4

=>x=2 hoặc x=-2

l:

(x-1)^4=16

=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

\(2^4.32>2^x>64\\ \Leftrightarrow2^4.2^5>2^x>2^6\\ \Leftrightarrow2^9>2^x>2^6\\ \Leftrightarrow9>x>6\)

=> x=7 hoặc x=8

2⁴.32>2^x>64

=>2⁹>2^x>2⁶

=>9>x>6

=>x€{8;7}

\(3\left(x+2\right)^2+\left(2x-3\right)^2-7\left(x-4\right)\left(x+4\right)=64\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-12x+9\right)-7\left(x^2-16\right)=64\)

\(\Leftrightarrow3x^2+12x+12+4x^2-12x+9-7x^2+112=64\)

\(\Leftrightarrow12+9+112=64\)(vô lí)

Vậy pt vô nghiệm

 TL:

\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-6x+9-7x^2+112=64\)

\(\Leftrightarrow6x+133=64\)  

\(\Leftrightarrow6x=-69\) 

\(\Leftrightarrow x=\frac{-23}{2}\) 

Vậy....

Nguyễn Văn Tuấn AnhSai từ dòng đầu tiên

\(\left(2x-3\right)^2=4x^2-12x+9\ne4x^2-6x+9\)