giúp mình nha mn

giúp mk với

16. 2 x  + 4. 2 x  = 5. 5 x  + 3. 5 x

⇔ 20. 2 x  = 8. 5 x  ⇔  2 / 5 x  =  2 / 5 1  ⇔ x = 1

2 x - 2 > 2 2 x + 1

⇔ |x−2| > 2|x+1|

⇔ x 2  − 4x + 4 > 4( x 2  + 2x + 1)

⇔ 3 x 2  + 12x < 0

⇔ −4 < x < 0

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2+6x+9}=2x+1\)

=>\(\left|x+3\right|=2x+1\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(2x+1\right)^2=\left(x+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(x-2\right)\left(3x+4\right)=0\end{matrix}\right.\Leftrightarrow x=2\)

\(\sqrt{x^2+6x+9}=2x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=2x-1\\ \Leftrightarrow\left|x+3\right|=2x-1\\ TH_1:x\ge-3\\ x+3=2x-1\Leftrightarrow-x=-4\Leftrightarrow x=4\left(tm\right)\\ TH_2:x< -3\\ -x-3=2x-1\Leftrightarrow-3x=2\Leftrightarrow x=-\dfrac{2}{3}\left(tm\right)\)

Vậy \(S=\left\{-\dfrac{2}{3};4\right\}\)

1a)mik ko biết là 2³⁰ hay 2³ + 0 . 3 nên mik làm cả hai cái

* 2³⁰ . 3 - (1¹⁰ + 8) : 3²

= 1073741824 . 3 - (1 + 8) : 9

= 1073741824 . 3 - 9 : 9

= 1073741824 . 3 - 1

= 3221225472 - 1

= 3221225471

* 2³ + 0 . 3 - (1¹⁰ + 8) : 3²

= 8 + 0 - (1 + 8) : 9

= 8 - 9 : 9

= 8 - 1

= 7

b) 4 . 5² - 3 x 2²

= 4 . 25 - 3 . 4

= 100 - 12

= 88

2) (105 - x) : 2⁵ = 3⁰ + 1

    (105 - x) : 32 = 1 + 1

    (105 - x) : 32 = 2

             105 - x = 2 . 32

             105 - x = 64

                      x = 105 - 64

             =>     x = 41

b) 5^x+1 = 125

    5^x+1 = 5³

       5^x = 5^3-1

       5^x = 5²

=> x = 2

c) 2x - 138 + 2² . 2²

= 2x - 138 + 2²⁺²

= 2x - 138 + 2⁴

= 2x - 138 + 16

= 2x - 154

x = 154 : 2

=> x = 77

#Học tốt!!!

\(\frac{x^2-4x+1}{x+1}+2=-\frac{x^2-5x+1}{2x+1}\):

\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)

PT \(\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}+1=0\)

         \(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{x+1}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)

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

\(\Leftrightarrow x=1;x=2;x=-\frac{2}{3}\)

Cả 3 giá trị trên đều thỏa mãn ĐKXĐ

Vậy PT đã cho có tập nghiêm : \(S=\left\{1;2;-\frac{2}{3}\right\}\)

Chúc bạn học tốt !!!

2 2 x - 2 . 2 x + 8 < 2 3 x . 2 1 - x ⇔ 2 2 x + 2 . 2 x - 8 > 0