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\(\left(64.27\right)^6.75^{18}=\left(2^6.3^3\right)^6.\left(3.5^2\right)^{18}=2^{36}.3^{36}.5^{36}=\left(2.3.5\right)^{36}=30^{36}\)
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a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
\(VP=18^{16}=2^{16}\cdot3^{32}\)
=> VT=VP
b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
(đề sai)
c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)
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Chia tất cả cho 30=>
\(\frac{x+y}{5}=\frac{y+z}{10}=\frac{x+z}{6}=\frac{x+z-x-y}{6-5}=\frac{y+z-x-z}{10-6}=\frac{z-y}{1}=\frac{y-x}{4}\)
\(\Leftrightarrow\frac{z-y}{3}=\frac{y-x}{12}\)
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\(\left(64.27\right)^6=4^{3.6}.3^{3.6}=\left(4.3\right)^{18}=12^{18}\)
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a) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=\dfrac{-3}{20}\)
vậy \(x=\dfrac{-3}{20}\)
b) \(\left|2x-1\right|=23\)
\(2x-1=\pm23\)
+) \(2x-1=23\Rightarrow2x=24\Rightarrow x=12\)
+) \(2x-1=-23\Rightarrow2x=-22\Rightarrow x=-11\)
vậy \(x\in\left\{-11;12\right\}\)
c) Xin sửa lại đề:
\(\sqrt{x} = 6\)
=> x = 62
x = 36
d) \(\frac{x}{4}=\frac{12}{18}\)
=> \(x = \frac{4.12}{18}=\frac{8}{3}\)
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a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
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a, \(\dfrac{-3}{4}.\dfrac{12}{-5}.\left(-\dfrac{25}{6}\right)=\dfrac{9}{5}.\left(-\dfrac{25}{6}\right)=-\dfrac{15}{2}\)
b,\(\left(-2\right).\dfrac{-38}{21}.\dfrac{-7}{4}.\left(-\dfrac{3}{8}\right)=\dfrac{76}{21}.\dfrac{-7}{4}.\left(-\dfrac{3}{8}\right)=-\dfrac{19}{3}.\left(-\dfrac{3}{8}\right)=\dfrac{19}{8}\)
c,\(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{4}{9}.\dfrac{3}{5}=\dfrac{4}{15}\)
d, \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]=\dfrac{7}{23}.\left(-\dfrac{13}{6}\right)=-\dfrac{91}{138}\)
like tớ với
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Thay f(17) và f(12) vào đa thức f(x)=ax+b ta có:
\(\hept{\begin{cases}12a+b=35\\17a+b=71\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=35-12a\\17a+35-12a=71\end{cases}}\)
\(\Leftrightarrow5a=36\)
\(\Leftrightarrow a=\frac{36}{5}\)
Theo đề bài \(a,b\in Z\)
Nên không thể đồng thời có f(17)=71 và f(12)=35
\(\left(64.27\right)^6=12^{18}\Leftrightarrow1728^6=12^8\)
Ta có :
\(12^{18}=\left(12^3\right)^6=1728^6\)
Vì \(1728^6=1728^6\)
\(\Leftrightarrow\left(64.27\right)^6=12^{18}\) đó là điều mà ta phải chứng minh