\(\left(8x-16\right)\left(x-5\right)=0\\ \Leftrightarrow8\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

(8\(x\) - 16).(\(x\) - 5) = 0

\(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)

 \(\left[{}\begin{matrix}8x=16\\x-5=0\end{matrix}\right.\)

  \(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy \(x\) \(\in\){2; 5}

\(\dfrac{x}{2}=\dfrac{x}{3}\Rightarrow x=0\)

\(a,A=2024=2^3\times11\times23\\B=8^5\times 125^6=\left(2^3\right)^5\times\left(5^3\right)^6=2^{15}\times5^{18}\\ b,Ư\left(84\right)=\left\{1;2;3;4;6;7;12;14;21;28;42;84\right\}\\\Rightarrow x\in\left\{1;2;3;4;6;7;12;14;21;28;42;84\right\}\\ x\in B\left(21\right)=\left\{0;21;42;63;84;105;126;147;168;189;210;....\right\}\)

\(\dfrac{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}\) + \(\dfrac{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}\) = 1 + 1 = 2

\(2x-124=x+224\\ \Rightarrow2x+x=224+124\\ \Rightarrow3x=348\\ \Rightarrow x=348:3\\ \Rightarrow x=116\)

2\(x\) - 124 = \(x\) + 224

2\(x\) - \(x\)    = 224 + 124

\(x\)           = 348

Đề này chưa đúng

\(\left(1^2+2^2+3^2+...+1000000^2\right).\left(91-273:3\right)\\ =\left(1^2+2^2+3^2+...+1000000^2\right).0=0\)

(1²+2²+3²+..+1000000²).(91-273:3)
=(1²+2²+3²+..+1000000²).(91-91)

=(1²+2²+3²+..+1000000²).0

=0

\(101ab=\left(100+1\right)ab\)

\(=100ab+ab\)

\(\overline{ab}\) \(\times\) 101 = \(\overline{ab}\) \(\times\)(100 + 1) = \(\overline{ab00}\) + \(\overline{ab}\) = \(\overline{abab}\) 

\(A=1+2+2^2+...+2^{2018}\)

\(2A=2+2^3+2^4+...+2^{2019}\)

\(A=2A-A=1-2^{2019}\)

\(B-A=2^{2019}-\left(1-2^{2019}\right)\)

\(B-A=2^{2019}-1+2^{2019}\)

\(B-A=1\)

`#3107`

\(A=1+2+2^2+2^3+...+2^{2018}\) và \(B=2^{2019}\)

Ta có:

\(A=1+2+2^2+2^3+...+2^{2018}\)

\(2A=2+2^2+2^3+...+2^{2019}\)

\(2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)

\(A=2+2^2+2^3+...+2^{2019}-1-2-2^2-2^3-...-2^{2018}\)

\(A=2^{2019}-1\)

Vậy, \(A=2^{2019}-1\)

Ta có:

\(B-A=2^{2019}-2^{2019}+1=1\)

Vậy, `B - A = 1.`

\(25\cdot32=25\cdot4\cdot8\)

\(=\left(25\cdot4\right)\cdot8=100\cdot8\)

\(=800\)

\(49\cdot101=49\cdot\left(100+1\right)\)

\(=49\cdot100+49\)

\(=4900+49\)

\(=4949\)