Question

Tìm x :

a) \(2^{4-x}\)= 32

b) ( x + 1,5 ) \(^2\)+ ( y - 2,5 ) \(^{10}\)= 0

c) 2\(^{-2}\). 2\(^x\)+ 2 .2\(^x\)= 9 . 2\(^6\)

d) 3\(^{-2}\). 3\(^4\). 3\(^x\)= 3\(^7\)

Answer 1

a, 2^4-x=32=2⁵

=> 4-x=5

<=> x=-1

b, (x+1,5)²+(y-2,5)¹⁰=0

Vì (x+1,5)2\(\ge\)0,   (y-2,5)¹⁰\(\ge\)0

\(\Rightarrow\hept{\begin{cases}x+1,5=0\\y-2,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=2,5\end{cases}}}\)

Answer 2

a)\(2^{4-x}\)=32

=>\(2^{4-x}\)=32=\(2^5\)

=>4-x=5

=>x=4-5=-1

=>x=-1

Answer 3

b) Ta có: \(\hept{\begin{cases}\left(x+1,5\right)^2\ge0;\forall x\\\left(y-2,5\right)^{10}\ge0;\forall x\end{cases}}\)

\(\Rightarrow\left(x+1,5\right)^2+\left(y-2,5\right)^{10}\ge0;\forall x\)

Do đó \(\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+1,5\right)^2=0\\\left(y-2,5\right)^{10}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x+1,5=0\\y-2,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=2,5\end{cases}}}\)

Vậy ...

Answer 4

a) 2^{4 - x} = 32

<=> 2^{4 - x} = 2⁵ 

<=> 4 - x = 5

<=> x = -1

=> x = -1

Answer 5

c) \(2^{-2}.2^x+2.2^x=9.2^6\) 

  \(2^x\left(2^{-2}+2\right)=9.2^6\) 

  \(2^x.\frac{9}{4}=9.2^6\) 

 \(2^x=4.2^6\) 

\(2^x=2^8\) 

\(\Rightarrow x=8\) 

d)\(3^{-2}.3^4.3^x=3^7\) 

  \(3^{-2+4+x}=3^7\)

 \(3^{2+x}=3^7\) 

\(\Rightarrow2+x=7\Leftrightarrow x=5\) 

Vậy.........