\(4^x-10.2^x+16=0\)

\(\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)

Đặt 2^x = t

\(\Rightarrow t^2-10t+16=0\)

\(\Leftrightarrow t^2-2t-8t+16=0\)

\(\Leftrightarrow t\left(t-2\right)-8\left(t-2\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t-8\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}t=2\\t=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2^x=2\\2^x=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(4^x-10\times2^x+16=0\)
\(\Leftrightarrow2^{2x}-2\times5\times2^x+16=0\)
\(\Leftrightarrow\left[\left(2^x\right)^2-2\times2^x\times5+25\right]-9=0\)
\(\Leftrightarrow\left(2^x-5\right)^2-3^2=0\)
\(\Leftrightarrow\left(2^x-5-3\right)\left(2^x-5+3\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-2\right)=0\)
\(\Leftrightarrow2^x-8=0\) hoặc \(2^x-2=0\)
\(\cdot2^x-8=0\Leftrightarrow2^x=8\Leftrightarrow x=3\)
\(\cdot2^x-2=0\Leftrightarrow2^x=2\Leftrightarrow x=1\)
Vậy \(S=\left\{3;1\right\}\)

\(4^x-10\cdot2^x+16=0\)

\(=\left(2^x\right)^2-10\cdot2^x+16=0\)

Đặt \(t=2^x\). Ta có:

\(t^2-10t+16=0\)

\(\Rightarrow t^2-2\cdot t\cdot5+25-9=0\)

\(\Rightarrow\left(t-5\right)^2-3^2=0\)

\(\Rightarrow\left(t-8\right)\left(t-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=8\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy S = {1,3}

\(4^x-10.2^x+16=0\)

\(\Leftrightarrow\left(2^x\right)^2-10.2^x=-16\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x\left(2^x-10\right)=2\left(2-10\right)\Leftrightarrow2^x=1\Leftrightarrow2^x=2^0\Leftrightarrow x=0\\2^x\left(2^x-10\right)=8\left(8-10\right)\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Leftrightarrow x=3\end{matrix}\right.\)

Vậy \(x\in\left\{0;3\right\}\)

a ) \(\left(3\times x-15\right)^7=0.\)

\(3\times x-15=0\)

\(3\times x=15\)

\(x=5\)

b ) \(10-\left\{\left[\left(x\div3+17\right)\div10+3\times2^4\right]\div10\right\}=5\)

\(10-\left\{\left[\left(x\div3+17\right)\div10+3\times16\right]\div10\right\}=5\)

\(10-\left\{\left[\left(x\div3+17\right)\div10+48\right]\div10\right\}=5\)

\(\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\left(x\div3+17\right)\div10+48=50\)

\(\left(x\div3+17\right)\div10=2\)

\(x\div3+17=20\)

\(x\div3=3\)

\(x=9\)

=1 đó bạn ạ

x = 1 nha !

\(2.2^2.2^3.2^4....2^x=1024=2^{10}\)

\(\Rightarrow2^{1+2+3+\text{4+}...+x}=2^{10}\)

\(\Rightarrow1+2+3+4+...+x=10\)

\(\Rightarrow1+2+3+4+...x=1+2+3+4\)=>x=4

a) (x - 1/2) x 2 = 9/16

=> x - 1/2 =  9/16 : 2

=> x - 1/2 = 9/16 x 1/2

=> x - 1/2 =9/32

=> x = 9/32 + 1/2

=> x = 25/32

b) |x + 1/2| = 3/4

=> x + 1/2 = 3/4 hoặc x + 1/2 =-3/4

=>x = 3/4 - 1/2 hoặc x = -3/4 -1/2

=>x = 1/4 hoặc x = -5/4

Vậy .........

2^x.2^{x + 1}.2^{x + 2} \(\le\) 1000...00 : 25⁹

<=> 2^{x + x + 1 + x + 2} \(\le\) 100⁹ : 25⁹

<=> 2^{3x + 3} \(\le\) 4⁹

<=> 2^{3x + 3} \(\le\) 2¹⁸

<=> 3x + 3 \(\le\) 18

<=> x \(\le\) 5

Mà x \(\in\) N => x \(\in\) {0; 1; 2; 3; 4; 5}

\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)

\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)

TA CÓ 1024=2^10

SUY RA X=0