\(x=\frac{10^2.5^3.15^6}{3^6.5^{10}}=\frac{\left(2.5\right)^2.\left(3.5\right)^6}{3^6.5^7}=\frac{2^2.5^2.3^6.5^6}{3^6.5^7}=2^2.5=20\)

Vậy x = 20

@Nguyễn Huy Tú bạn có thể viết rõ cách làm có dc ko?

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(5^{x+3}\left(5-3\right)=2.5^{11}\)

\(5^{x+3}.2=2.5^{11}\)

\(5^{x+3}=5^{11}\)

\(x+3=11\)

\(x=8\)

\(4^{x+3}-3.4^{x+1}=13.4^{11}\)

\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)

\(4^{x+1}.13=13.4^{11}\)

\(4^{x+1}=4^{11}\)

\(x+1=11\)

\(x=10\)

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

Mình làm tiếp câu b nha !

b,                                               Bài giải

\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)

\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)

\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)

\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)

\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)

\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)

\(x-2016=-2\text{ : }\frac{31}{56}\)

\(x-2016=-\frac{112}{31}\)

\(x=-\frac{112}{31}+2016\)

\(x=\frac{62384}{31}\)

Tìm số ự nhiên x biết.

a, 2^x=4^6.16³

2^x=(2²)².(2⁴)³

2^x=2⁴.2¹²

2^x=2¹⁶

=>x=16

Vậy x=16

học tốt

a, \(2^x=4^6\cdot16^3\)

\(2^x=\left(2^2\right)^6\cdot\left(2^4\right)^3\)

\(2^x=2^{12}\cdot2^{12}\)

\(2^x=2^{24}\)

\(\Rightarrow\text{ }x=24\)

\(b,\text{ }\frac{2^{x-3}}{4^{10}}=8^3\)

\(\Rightarrow\text{ }\frac{2^{x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\)

\(\frac{2^{x-3}}{2^{20}}=2^9\)

\(\Rightarrow\text{ }2^{x-3}=2^{20}\cdot2^9\)

\(2^{x-3}=2^{29}\)

\(\Rightarrow\text{ }x-3=29\)

\(x=29+3\)

\(x=32\)

a) \(\left|x-\frac{1}{2}\right|+3=2^2\)

\(\Leftrightarrow\left|x-\frac{1}{2}\right|+3=4\)\(\Leftrightarrow\left|x-\frac{1}{2}\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=-1\\x-\frac{1}{2}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}\)

Vậy \(x=-\frac{1}{2}\)hoặc \(x=\frac{3}{2}\)

b) \(3^x+3^{x+2}=10^2-2.5\)

\(\Leftrightarrow3^x+3^x.3^2=100-10\)

\(\Leftrightarrow3^x+3^x.9=90\)

\(\Leftrightarrow3^x.\left(1+9\right)=90\)

\(\Leftrightarrow3^x.10=90\)

\(\Leftrightarrow3^x=9=3^2\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

c) \(3-2x^2=\frac{5}{2}\)

\(\Leftrightarrow2x^2=3-\frac{5}{2}\)\(\Leftrightarrow2x^2=\frac{1}{2}\)

\(\Leftrightarrow x^2=\frac{1}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{2}\end{cases}}\)

Vậy \(x=-\frac{1}{2}\)hoặc \(x=\frac{1}{2}\)

theo mik thì câu a chỉ cần bằng 3/2 là được bạn ah

\(\left|x\right|-3=4\)

\(\Rightarrow\left|x\right|=4+3\)

\(\Rightarrow\left|x\right|=7\)

\(\Rightarrow x\in\left\{-7;7\right\}\)

\(\frac{10^3+2.5^3+5^3}{55}=\frac{2^3.5^3+5^3\left(1+2\right)}{5.11}=\frac{8.5^3+5^3.3}{5.11}=\frac{5^3\left(8+3\right)}{5.11}\)

\(=\frac{5^3.11}{5.11}=\frac{5^3}{5}=5^2=25\)

Bài 1 :

\(\left|x\right|-3=4\)

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

Vậy...

Bài 2 :

\(\frac{10^3+2\cdot5^3+5^3}{55}\)

\(=\frac{2^3\cdot5^3+2\cdot5^3+5^3}{5\cdot11}\)

\(=\frac{5^3\cdot\left(2^3+2+1\right)}{5\cdot11}\)

\(=\frac{5^3\cdot11}{5\cdot11}\)

\(=5^2\)