a, \(x^2=16\)

\(\Rightarrow x^2=4^2\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)

c, \(\frac{175}{5^{x+1}}=7\)

\(\Rightarrow5^{x+1}=175:7\)

\(\Rightarrow5^{x+1}=25\)

\(\Rightarrow5^{x+1}=5^2\Rightarrow x+1=2\Rightarrow x=1\)

a, \(x^2=\dfrac{1}{16}\Rightarrow x=\pm\dfrac{1}{4}\)

b, \(x^5:x^2=-\dfrac{1}{64}\Rightarrow x^3=\left(-\dfrac{1}{4}\right)^3\Rightarrow x=-\dfrac{1}{4}\)

c, \(x^3:x^2=\dfrac{32}{243}\Rightarrow x=\dfrac{32}{243}\)

d, \(\left(x^2\right)^2=\dfrac{81}{16}\Rightarrow x^4=\left(\dfrac{3}{2}\right)^4\Rightarrow x=\pm\dfrac{3}{2}\)

Chúc bạn học tốt!!!

3) Tìm x

a) \(^{x^2}\)=\(\dfrac{1}{16}\)

<=> x = \(\sqrt{-\dfrac{1}{16}}\)

\(\sqrt{\dfrac{1}{16}}\)

<=> x = -14

+14

b) \(x^{5^{ }}\): \(x^2\) = \(-\dfrac{1}{64}\)

<=> \(^{x^{5-2}}\) =\(-\dfrac{1}{64}\)

<=> \(x^3\) = \(-\dfrac{1}{64}\)

<=> x = \(-\dfrac{1}{4}\)

c)\(x^3:x^2\) = \(\dfrac{32}{243}\)

<=> \(^{x^{3-2}}\) = \(\dfrac{32}{243}\)

<=> x = \(\dfrac{32}{243}\)

d) \((x^2)^2\) = \(\dfrac{81}{16}\)

<=>\(^{x^{2.2}}\) = \(\dfrac{81}{16}\)

<=> \(x^4\) = \(\dfrac{81}{16}\)

<=> x = \(\dfrac{3}{2}\)

\(-\dfrac{3}{2}\)

16/2^x = 2

=> 2^x . 2 = 16

     2^x      = 8

     2^3     = 2^3

=> x = 3

\(\frac{16}{2^x}=2\Rightarrow16:2^x=2\)

\(\Rightarrow2^x=16:2=8\Rightarrow2^x=2^3\Rightarrow x=3\)

\(\frac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x:81=-27\)

\(\Rightarrow\left(-3\right)^x=-27\cdot81=-2187\Rightarrow\left(-3\right)^x=\left(-3\right)^7\Rightarrow x=7\)

mk biết mỗi thế thôi. mong bạn thông cảm.

~CÁC BẠN GIÚP MK LÊN 200 ĐIỂM NHA. BẠN NÀO GIÚP THÌ MK K LẠI NHÉ.~

~THANKS~

Để đẳng thức trên xảy ra thì một trong các thừa số của tích phải bằng 0.

- Xét x - 4 =0 <=> x = 4

- Xét x² + 16 = 0 <=> x² = - 16 (loại vì x² > 0)

- Xét x² - 16 = 0 <=> x² = 16 <=> x = + 4

- Xét x + 1 = 0 <=> x = -1

Vật x \(\in\) {-1;+4}

1/ 2^x = 4⁵.4⁶

=> 2^x = 4^{5 + 6}

=> 2^x = 4¹¹

=> 2^x = (2²)¹¹

=> 2^x = 2²²

=> x = 22

vậy_

2/ 2^x = 4⁶.16³

=> 2^x = (2²)⁶.(2⁴)³

=> 2^x = 2¹².2¹²

=> 2^x = 2^{12 + 12}

=> 2^x = 2²⁴

=> x = 24

3/ 2^x = 4⁵.16²

=> 2^x = (2²)⁵.(2⁴)²

=> 2^x = 2¹⁰.2⁸

=> 2^x = 2^{10 + 8}

=> 2^x = 2¹⁸

=> x = 18

vậy_

\(\frac{1}{2^x}=4^5.4^3=4^{5+3}=4^8\)

\(\Rightarrow1=4^8.2^x=2^{2.8+x}=2^{16+x}\)

ta có 1 < 2¹ =>  2^16+x < 2¹

=> 2^16+x = 2⁰

=> 16+x=0

=> x= -16

a, |x-7|=|9-234|

=> |x-7|=|-225|

=> |x-7|=225

=>\(\orbr{\begin{cases}x-7=225\\x-7=-225\end{cases}}\)=>\(\orbr{\begin{cases}x=232\\x=-218\end{cases}}\)

b, \(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

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

=>\(\orbr{\begin{cases}\frac{x}{8}=\frac{1}{2}\\\frac{x}{8}=\frac{-1}{2}\end{cases}}\)=>\(\orbr{\begin{cases}x.2=8\\x.2=-8\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

c, (8-3²)²+(x-39)²=10

=>(-1)²+(x-39)²=10

=>(x-39)²=10-1

=>(x-39)²=9

=>(x-39)²=3²

=>\(\orbr{\begin{cases}x-39=3\\x-39=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=42\\x=36\end{cases}}\)

a)\(\left|x-7\right|=\left|9-234\right|\)

\(\Rightarrow\left|x-7\right|=\left|-225\right|\)

\(\Rightarrow\left|x-7\right|=225\)

\(\Rightarrow x-7=\pm225\)

• \(x-7=225\Rightarrow x=232\)
• \(x-7=-225\Rightarrow x=-218\)

Vậy \(x\in\left\{232;-218\right\}\)

b)\(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{4}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\left(\pm\frac{1}{2}\right)^2\)

\(\Rightarrow\frac{x}{8}=\pm\frac{1}{2}\)

• \(\Rightarrow\frac{x}{8}=\frac{1}{2}\Rightarrow\frac{x}{8}=\frac{4}{8}\Rightarrow x=4\)
• \(\Rightarrow\frac{x}{8}=\frac{-1}{2}\Rightarrow\frac{x}{8}=\frac{-4}{8}\Rightarrow x=-4\)

Vậy \(x\in\left\{\pm4\right\}\)

c)\(\left(8-3^2\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(8-9\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(-1\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow1+\left(x-39\right)^2=10\)

\(\Rightarrow\left(x-39\right)^2=9\)

\(\Rightarrow\left(x-39\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow x-39=\pm3\)

• \(\Rightarrow x-39=3\Rightarrow x=42\)
• \(\Rightarrow x-39=-3\Rightarrow x=36\)

Vậy \(x\in\left\{42;36\right\}\)

\(a,16^x:4^x=16\)

\(\left(4^x\right)^2:4^x=4^2\)

\(\Rightarrow4^x=4^2\Leftrightarrow x=2\)

\(b,2^{-1}.2^x+4.2^x=72\)

\(\Rightarrow2^{x-1}+2^{x+2}=72\)

\(\Rightarrow2^{x-1}\left(1+2^3\right)=72\)

\(\Rightarrow2^{x-1}=72:9=8=2^3\)

\(\Rightarrow x=4\)

\(c,\left(2^x+1\right)^3=-64\)

\(\Rightarrow2^x+1=-4\)

\(\Rightarrow2^x=-5\)

a) 16^x : 4^x = 16 

<=> ( 16 : 4 )^x = 16 

<=> 4^x = 16

<=> 4^x = 4^2 

=> x = 2

Vậy x =2

b) 3.[5^2x+1 - (5²)^x] = 3.100

3( 5^2x+1-5^2x) = 3.100

-> 5^2x+1-5^2x = 100

mà 5³-5²= 100

-> 5^2x+1-5^2x = 5³-5²

^{Vậy x = 1}

c) (4²)^x : 4^x = 4²

<=>4^2x-x =4²

<=> 4^x = 4²

-> x=2

d) 2^x( 0,5 +4) = 72

2^x. 4,5 = 72

2^x = 72:4,5 = 16

2^x = 2⁴ = 16

-> x= 4

Mình sẽ giúp bạn giải từng phương trình nhé:

1. \(x^{3} = - 27\)

Lấy căn bậc 3 hai vế:

\(x = \sqrt[3]{- 27} = - 3\)

✅ Nghiệm: \(x = - 3\)

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

Lấy căn bậc 3 hai vế:

\(2 x - 1 = \sqrt[3]{8} = 2\)

Giải tiếp:

\(2 x = 3 \Rightarrow x = \frac{3}{2}\)

✅ Nghiệm: \(x = \frac{3}{2}\)

3. \(\left(\right. x - 2 \left.\right)^{2} = 16\)

Lấy căn bậc 2 hai vế (lưu ý ±):

\(x - 2 = \pm 4\)

Giải:

• \(x - 2 = 4 \Rightarrow x = 6\)
• \(x - 2 = - 4 \Rightarrow x = - 2\)

✅ Nghiệm: \(x = 6\) hoặc \(x = - 2\)

📌 Tóm tắt các nghiệm:

1. \(x = - 3\)
2. \(x = \frac{3}{2}\)
3. \(x = 6\) hoặc \(x = - 2\)

Nếu bạn cần vẽ đồ thị hoặc kiểm tra lại bằng máy tính, cứ nói nhé!

tk