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a.

\(\frac{2^{10}\times13+2^{10}\times65}{2^8\times104}=\frac{2^{10}\times\left(13+65\right)}{2^8\times104}=\frac{2^2\times78}{104}=\frac{4\times78}{104}=\frac{312}{104}=3\)

b.

\(\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(65\times111-15\times37\times13\right)\)

\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(7215-7215\right)\)

\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times0\)

= 0

a. 3

b. 0

Bài 2:

a) \(\left(x-3\right)^3+27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0-27\)

\(\Leftrightarrow\left(x-3\right)^3=-27\)

\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=\left(-3\right)+3\)

\(\Leftrightarrow x=0\)

b) \(-125-\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=-125-0\)

\(\Leftrightarrow\left(x+1\right)^3=-125\)

\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=\left(-5\right)-1\)

\(\Leftrightarrow x=-6\)

c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)

\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}:2\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

d) \(2^x+2^{x+1}=24\)

\(\Leftrightarrow2^x+2^x.2=24\)

\(\Leftrightarrow2^x\left(1+2\right)=24\)

\(\Leftrightarrow2^x.3=24\)

\(\Leftrightarrow2^x=24:3\)

\(\Leftrightarrow2^x=8\)

\(\Leftrightarrow2^x=2^3\)

\(\Rightarrow x=3\)

e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)

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

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

g) \(\left|x-3\right|+2x=10\)

\(\Leftrightarrow\left|x-3\right|=10-2x\)

\(\Leftrightarrow\left|x-3\right|=2.5-2x\)

\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)

(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)

Bài 1:

a) \(2^7+2^9⋮10\)

Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)

\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)

\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)

\(\Leftrightarrow\overline{C8}+\overline{D2}\)

\(\Leftrightarrow\overline{E0}\)

Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)

b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)

Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)

\(\Leftrightarrow2^{72}.5^{20}\)

Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)

c) \(3^{10}+3^{12}⋮30\)

Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)

\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)

\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)

\(\Leftrightarrow\overline{C9}+\overline{B1}\)

\(\Leftrightarrow\overline{D0}⋮10\)

(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)

a,\(x^4:9=9\\ \Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

b,\(x^2-30=6\\ \Leftrightarrow x^2=36\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

c,\(x^2:10=10\Leftrightarrow x^2=\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

x⁴ : 9 = 9

x⁴ = 9 . 9

x⁴ = 81

x⁴ = 3⁴

=> x = 3

x² - 30 = 6

x² = 6 + 30

x² = 36

x² = 6²

=> x = 6

x² : 10 = 10

x² = 10 . 10

x² = 100

x² = 10²

=> x = 10

mình ko hỉu bạn nói gì cho lắm ＼（〇_ｏ）／

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}=\dfrac{\left(y^2-x^2\right)-\left(y^2+x^2\right)}{3+5}=\dfrac{\left(y^2-x^2\right)-\left(y^2-x^2\right)}{3-5}\Rightarrow\dfrac{2y^2}{8}=\dfrac{-2x^2}{-2}\Rightarrow\dfrac{y^2}{4}=x^2\Rightarrow y^2=4x^2\)

Ta có: \(x^{10}.y^{10}=x^{10}.\left(4x^2\right)^5=1024.x^{20}=1024\Rightarrow x^{20}=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\Rightarrow y^2=4\Rightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1\right\}\) và \(y\in\left\{4;-4\right\}\)

\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}\)

\(\Leftrightarrow5\left(y^2-x^2\right)=3\left(y^2+x^2\right)\)

\(\Leftrightarrow5y^2-5x^2=3y^2+3x^2\)

\(\Leftrightarrow2y^2=8x^2\)

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

\(\Leftrightarrow y^{10}=1024.x^{10}\)

Mà \(x^{10}.y^{10}=1024\)

\(\Leftrightarrow x^{10}.1024x^{10}=1024\)

\(\Leftrightarrow x^{20}=1\)

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

+)Với \(x=1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

+) Với \(x=-1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy...