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A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=2.2.2.2.2
A=32
\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)
\(2\cdot2\cdot2\cdot2\cdot2=2^5\)
\(=32\)
\(\frac{3^6.45^4-15^{13}.9^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3.3.5\right)^4-\left(3.5\right)^{13}-\left(3^2\right)^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3.15\right)^6}=\frac{3^6.3^4.3^4.5^4-3^{13}.5^{13}-3^{-18}}{3^{12}.5^6+3^6.15^6}=\frac{3^{14}.5^4-3^{13}.5^{13}-3^{-18}}{3^{12}.5^6+3^6.\left(3.5\right)^6}=\frac{ }{ }\)
Mình nghĩ hết được rồi
giúp mk với
\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(5^{-1}\cdot25^x=125\)
Lưu ý: dấu'.' là dấu nhân
Ta có : (x - 1)2 = (x - 1)4
=> (x - 1)4 - (x - 1)2 = 0
=> (x - 1)2.[(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 0 => x = 1
Nếu x - 1 = 1 => x = 2
Nếu x - 1 = - 1 => x = 0
Vậy \(x\in\left\{0;1;2\right\}\)
b) 5 - 1 . 25x = 125
=> \(\frac{1}{5}.25^x=125\)
=> 25x = 625
=> 25x = 252
=> x = 2
Vậy x = 2
a) \(\left(x-1\right)^2=\left(x-1\right)^4\Leftrightarrow1=\left(x-1\right)^2\)\(\Leftrightarrow x-1=1\Leftrightarrow x=2\)
b) \(5^{-1}.25^x=125\Leftrightarrow5.25^{x-1}=125\Leftrightarrow25^{x-1}=25\)\(\Rightarrow x-1=1\Leftrightarrow x=2\)
\(=\frac{2^{2018}.9^{2018}.9^{2018}}{9^{4036}.2^{2016}}\)
\(=\frac{2^{2018}.9^{4036}}{9^{4036}.2^{2016}}\)
\(=2^2\)
\(=4\)
\(a,\frac{20^{12}\cdot6^{14}}{8^{13}\cdot15^{12}}\)
\(=\frac{5^{12}\cdot2^{24}\cdot2^{14}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}\)
\(=\frac{5^{12}\cdot2^{38}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}=\frac{3^2}{2}=\frac{9}{2}\)
\(b,\frac{45^{12}\cdot10^{14}}{18^{13}\cdot25^{12}}\)
\(=\frac{5^{12}\cdot3^{24}\cdot2^{14}\cdot5^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}\)
\(=\frac{5^{26}\cdot3^{24}\cdot2^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}=\frac{5^2\cdot2}{3^2}=\frac{50}{9}\)
\(c,\frac{18^{12}\cdot27^8}{6^8\cdot3^{40}}\)
\(=\frac{2^{12}\cdot3^{24}\cdot3^{24}}{2^8\cdot3^8\cdot3^{40}}\)
\(=\frac{2^{12}\cdot3^{48}}{2^8\cdot3^{48}}=2^4=16\)
\(d,\frac{12^{14}\cdot9^{18}}{8^9\cdot27^{17}}\)
\(=\frac{3^{14}\cdot2^{28}\cdot3^{36}}{2^{27}\cdot3^{51}}\)
\(=\frac{3^{50}\cdot2^{28}}{2^{27}\cdot3^{51}}=\frac{2}{3}\)
làm hơi tắt nên chịu khó hiểu
Ta có: x + y + z = 36 . (2018 - 2019) = 36 . (-1) = -36
Lại có: \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)\(\Rightarrow\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4x\right)}{9}=\frac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)
Do đó: \(\frac{3x-2y}{4}=0\)\(\Rightarrow3x-2y=0\)\(\Rightarrow3x=2y\)\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)(1)
\(\frac{2z-4x}{3}=0\)\(\Rightarrow2z-4x=0\)\(\Rightarrow2z=4x\)\(\Rightarrow\frac{x}{2}=\frac{z}{4}\)(2)
Từ (1), (2) \(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{-36}{9}=-4\)
Do đó: \(\hept{\begin{cases}\frac{x}{2}=-4\\\frac{y}{3}=-4\\\frac{z}{4}=-4\end{cases}\Rightarrow}\hept{\begin{cases}x=-8\\y=-12\\z=-14\end{cases}}\)
Vậy...
\(\frac{5^{102}\cdot9^{1000}}{3^{2018}\cdot25^{50}}=\frac{5^{102}\cdot3^{2000}}{3^{2018}\cdot5^{100}}=\frac{5^2}{3^{18}}\)