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a) Cách 1: \(\left(3^2\right)^3=3^{2.3}=3^6\)
\(\left(3^3\right)^2=3^{3.2}=3^6\)
\(\left(3^2\right)^5=3^{2.5}=3^{10}\)
\(9^8=\left(3^2\right)^8=3^{2.8}=3^{16}\)
\(27^6=\left(3^3\right)^6=3^{3.6}=3^{18}\)
\(81^{10}=\left(3^4\right)^{10}=3^{4.10}=3^{40}\)
Cách 2: \(\left(3^2\right)^3=9^3\)
\(\left(3^3\right)^2=3^{3.2}=\left(3^2\right)^3=9^3\)
\(\left(3^2\right)^5=9^5\)
\(9^8\)
\(27^6=\left(3^3\right)^6=3^{3.6}=3^{18}=3^{2.9}=\left(3^2\right)^9=9^9\)
\(81^{10}=\left(9^2\right)^{10}=9^{2.10}=9^{20}\)
Trả lời :
b)
Ta có : \(5^{28}=5^{2.14}=\left(5^2\right)^{14}=25^{14}< 26^{14}\)
\(\Rightarrow5^{28}< 26^{14}\)
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
\(a,\) \(\left(3^2\right)^3\) = \(3^{2.3}\) = \(3^6\)
\(\left(3^3\right)^2\) = \(3^{3.2}=3^6\)
\(\left(3^2\right)^5\) = \(3^{2.5}=3^{10}\)
\(9^8=\left(3^2\right)^8=3^{2.8}=3^{16}\)
b, \(\left(5^3\right)^2=5^{3.2}=5^6\)
\(\left(5^4\right)^3=5^{4.3}=5^{12}\)
\(\left(5^2\right)^4\) = \(5^{2.4}=5^8\)
\(25^5=\left(5^2\right)^5=5^{2.5}=5^{10}\)
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
\(\left(A\right)125^{80}và25^{118}\)
\(125^{80}=\left(5^3\right)^{80}=5^{3.80}=5^{240}\)
\(25^{118}=\left(5^2\right)^{118}=5^{2.118}=5^{236}\)
Vì \(5^{240}>5^{236}\)nên \(125^{80}>25^{118}\)
\(\left(B\right)4^{21}và64^7\)
\(4^{21}\)giữ nguyên
\(64^7=\left(4^3\right)^7=4^{3.7}=4^{21}\)
Vì \(4^{21}=4^{21}\)nên \(4^{21}=64^7\)
dễ mà bạn,mình chưa học mà mình biết rồi nè.
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
Câu 1: ta có:
\(4C=4^2+4^3+...+4^n+4^{n+1}\)lấy 4C-C ta có:\(3C=4^{n+1}-4\)
=> C=\(\frac{4^{n+1}-4}{3}\)
b, tương tự ta có: \(5D=5+5^2+...+5^{2000}+5^{2001}\)
=> D=\(\frac{5^{2001}-1}{4}\)
Câu 2: ta có: \(2A=2+2^2+2^3+...+2^{200}+2^{201}\)
=> Lấy 2A - A, ta có: \(A=2^{201}-1\)=> A+1=2201 -1+1=2201 .
Vậy \(A+1=2^{201}\)
Câu 3: Ta có: \(3B=3^2+3^3+3^4+...+3^{2005}+3^{2006}\)
=> \(B=\frac{3^{2006}-3}{2}\)=> \(2B+3=3^{2006}-3+3=3^{2006}\)
Vậy 2B + 3 là một lũy thừa của 3...
Câu 4: Do 4=22nên ta có: \(2C=2^3+2^3+2^4+...+2^{2005}+2^{2006}\)
=> \(C=2^{2006}+2^3-\left(2^2+4\right)\)=>\(C=2^{2006}\)
Vậy C là lũy thừa của 2 có số mũ là 2006
Câu 5: a, Do 3n+2 chia hết cho n-1 hay:
3n-3+5 sẽ chia hết cho n-1 =>3(n-1) +5 chia hết cho n-1...mà 3(n-1) chia hết cho n-1 nên 5 chia hết n-1;
=> n-1 thuộc (1,5,-1,-5);;; nên n tương ứng với(2;6;0;-4)
b ,Do n+6 chia hết cho n nên 6 chia hết cho n hay n là ước của 6
nên => n thuộc (1,6,-1,-6);
c, Do 3n+4 chia hết cho n-1 hay: 3n-3+7 chia hết cho n-1
=> 3(n-1)+7 chia hết cho n-1 => 7 chia hết cho n-1;
n -1 thuộc (1,7,-1,-7) hay n sẽ tương ứng với( 2,8,0,-6);
d, Do n+5 chia hết cho n+1 hay n+1+4 chia hết cho n+1
=> 4 chia hết cho n+1 => n+1 thuộc (1,4,-1,-4) nên n tương ứng với (0,3,-2,-5);
182=22.34
103=23.53.vì 3>2=>23>22
53=125 mà 34=81=> 5^3>3^4.vậy....
a, \(18^2=324\)
\(10^3=1000\)
Vì \(1000>324\Rightarrow10^3>18^2\)
b, \(3^2+4^2=9+16=25\)
\(\left(3+4\right)^2=7^2=49\)
Vì \(49>25\Rightarrow\left(3+4\right)^2>3^2+4^2\)