\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)

\(\Leftrightarrow2^{x-1}=24-16+3-3\)

\(\Leftrightarrow x-1=3\)

hay x=4

\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)

<=> \(3+2^{x-1}=11\)

<=> \(2^{x-1}=8\)

<=> \(2^{x-1}=2^3\)

<=> x - 1 = 3

<=> x = 4

`3 + 2^(x-1)=24-[4^2-(2^2-1)]`

`=> 3+2^(x-1)=24-[16-(4-1)]`

`=>3+2^(x-1)=24-[16-3]`

`=> 3+2^(x-1)=24-13`

`=> 3+2^(x-1)=11`

`=> 2^(x-1)=11-3`

`=> 2^(x-1)=8`

`=> 2^(x-1)=2^3`

`=> x-1=3`

`=> x=3+1`

`=> x=4`

Vậy `x=4`

\(a,\Rightarrow2^{x-1}=24-\left(16-3\right)-3\\ \Rightarrow2^{x-1}=24-13-3\\ \Rightarrow2^{x-1}=8=2^3\\ \Rightarrow x-1=3\Rightarrow x=4\\ b,\Rightarrow\left(19x+50\right):14=25-16=9\\ \Rightarrow19x+50=126\\ \Rightarrow x=4\)

3 + 2x - 1= 24 - [42 - (22 - 1)

3 + 2x - 1= 24 - [42 - 21]

3 + 2x - 1= 24 - 21

3 + 2x - 1= 3

3 + 2x = 3 + 1

3 + 2x = 4

2x = 4 - 3

2x =1

x = 1:2

x = 0,5

Vậy x = 0,5

3 + 2x - 1= 24 - [42 - (22 - 1)

3 + 2x - 1= 24 - [42 - 21]

3 + 2x - 1= 24 - 21

3 + 2x - 1= 3

3 + 2x = 3 + 1

3 + 2x = 4

2x = 4 - 3

2x =1

x = 1:2

x = 0,5

suy ra x = 0,5

3+2x-1=24-[42-(22-1)]

3+2x-1=24-(42-21)

3+2x-1=24-21

3+2x-1=3

3+2x=3+1

3+2x=4

2x=4-3

2x=1

x=1:2

x=0,5

1) \(\left(x+1\right)^2=x^2+2x+1\)

2) \(\left(2x+1\right)^2=4x^2+4x+1\)

3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)

4) \(\left(2x+3\right)^2=4x^2+12x+9\)

5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)

6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)

7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)

8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)

9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)

10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)

3: =>x(x+1)=0

=>x=0 hoặc x=-1

4: =>(2x-3)(x+2)=0

=>x=3/2 hoặc x=-2

6: =>6x=7 hoặc 6x=-7

=>x=7/6 hoặc x==7/6

1: \(=\dfrac{x-1}{x^2+x+1}+\dfrac{x+1}{x-1}\)

\(=\dfrac{x^2-2x+1+x^3+x^2+x^2+x+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^3+3x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}\)

2: \(=\dfrac{\left(x^2-y^2\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)^2}{x^2+xy+y^2}\)

a: \(2\left(x-51\right)=2\cdot2^3+20\)

=>\(2\left(x-51\right)=2^4+20=36\)

=>x-51=36/2=18

=>x=18+51=69

b: \(2x-49=5\cdot3^2\)

=>\(2x-49=5\cdot9=45\)

=>2x=45+49=94

=>x=94/2=47

c: \(\left[\left(8x-12\right):4\right]\cdot3^3=3^6\)

=>\(\left[4\cdot\dfrac{\left(2x-3\right)}{4}\right]=3^3\)

=>\(2x-3=3^3=27\)

=>2x=3+27=30

=>x=30/2=15

d: \(2^{x+1}-2^2=32\)

=>\(2^{x+1}=32+2^2=32+4=36\)

=>\(x+1=log_236\)

=>\(x=log_236-1\)

e: \(\left(x^3-77\right):4=5\)

=>\(x^3-77=20\)

=>\(x^3=77+20=97\)

=>\(x=\sqrt[3]{97}\)

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

\(\Leftrightarrow x^3-x^3-1=x\)

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)

=>\(5^{x-2}=16+9=25\)

=>x-2=2

=>x=4

2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)

=>3^x=0

=>x=0

3: \(\Leftrightarrow2^x+2^x\cdot16=272\)

=>2^x*17=272

=>2^x=16

=>x=4

4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)

=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)

=>2^x-1=8

=>x-1=3

=>x=4