\(2^x-26=6\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\)

\(2^x-26=6\)

⇒ \(2^x=6+26=32\)

⇒ \(2^x=32=2^5\)

⇒ \(x=5\)

\(2^x-26=6\)

\(2^x=6+26=32\)

\(2^x=2^5\)

=> x = 5

Vậy x = 5

Vì 1⁵= 1⁶ nên x - 5 = 1

x= 1+5

x=6

\(\left(x-5\right)^5=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)

\(\Rightarrow\left(x-5\right)^5\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(\left(x-5\right)^5=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)

\(\Rightarrow\left(x-5\right)^5.\left(x-5-1\right)=0\)

\(\Rightarrow\left(x-5\right)^5.\left(x-6\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^5=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(\sqrt{x+1}+\sqrt{4x+4}=6\)

\(\Leftrightarrow\sqrt{x+1}+4\sqrt{x+1}=6\)

\(\Leftrightarrow5\sqrt{x+1}=6\)

\(\Leftrightarrow\sqrt{x+1}=\dfrac{6}{5}\)

\(\Leftrightarrow x+1=\left(\dfrac{6}{5}\right)^2\)

\(\Leftrightarrow x+1=\dfrac{36}{25}\)

`<=> x= 11/25`

** Bổ sung điều kiện $x,y$ là các số nguyên. 

$x+5y+xy=6$

$(x+xy)+5y=6$

$x(1+y)+5(y+1)=11$

$(y+1)(x+5)=11$
Vì $x,y$ nguyên nên $x+5, y+1$ cũng nguyên. Ta xét các TH sau:

TH1: $x+5=1, y+1=11\Rightarrow x=-4; y=10$
TH2: $x+5=11, y+1=1\Rightarrow x=6; y=0$

TH3: $x+5=-1; y+1=-11\Rightarrow x=-6; y=-12$

TH4: $x+5=-11; y+1=-1\Rightarrow x=-16; y=-2$

=41/10

x - 1/2 = 18/5

x = 18/5 + 1/2

x = 41/10

\(x-\dfrac{1}{2}=\dfrac{3}{5}\times6\)

\(x-\dfrac{1}{2}=\dfrac{18}{5}\)

\(x=\dfrac{18}{5}+\dfrac{1}{2}\)

\(x=\dfrac{41}{10}\)

\(P=\dfrac{6}{x}+\dfrac{3}{2}x+\dfrac{24}{y}+\dfrac{3}{2}y-\dfrac{1}{2}\left(x+y\right)\ge2\sqrt{6.\dfrac{3}{2}}+2\sqrt{24.\dfrac{3}{2}}-\dfrac{1}{2}.6=15\Rightarrow min=15\Leftrightarrow x=2;y=4\)

\(B=x+y+\dfrac{6}{x}+\dfrac{24}{y}=\left(\dfrac{3x}{2}+\dfrac{6}{x}\right)+\left(\dfrac{3y}{2}+\dfrac{24}{y}\right)-\dfrac{3}{2}\left(x+y\right)\)

\(B\ge2\sqrt{\dfrac{18x}{2x}}+2\sqrt{\dfrac{72y}{2y}}-\dfrac{3}{2}.6=15\)

\(B_{min}=15\) khi \(\left(x;y\right)=\left(2;4\right)\)

\(\dfrac{6}{x}+8y\) hay \(\dfrac{6}{x}+\dfrac{8}{y}\)? Nếu là 8y tại sao ko cộng luôn với 2y thành 10y nhỉ?

đề ra thế 

cái tứ 1

tất cả số nguyên x >0 là được

Để B>0 thì x+2>0

hay x>-2