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\(5^{36}\)và \(11^{24}\)
Ta có :
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vi \(125>121\)nên \(5^{36}>11^{24}\)
\(3^{100}\)và \(2^{150}\)
Ta có :
\(3^{100}=\left(3^2\right)^{50}=9^{50}\)
\(2^{150}=\left(2^3\right)^{50}=8^{50}\)
Vì \(9>8\)nên \(3^{100}>2^{150}\)

TH1 : \(16a7b⋮2;5\Rightarrow b=0\)
\(để 16a70⋮3\Rightarrow1+6+a+7 ⋮3\)
\(\Rightarrow a\in\left\{1;4;7\right\}\)
TH2 : \(3a86b⋮2;5\Rightarrow b=0\)
\(để3a860⋮9\Rightarrow3+a+8+6⋮9\)
\(\Rightarrow a=1\)
TH3 : 53a7b : 5 dư 2
\(\Rightarrow b\in2;7\)
\(để53a72⋮3\Rightarrow5+3+a+7+2⋮3\)
\(\Rightarrow a\in1;4;7\)
\(để53a77⋮3\Rightarrow5+3+a+7+7⋮3\)
\(\Rightarrow a\in2;5;8\)
TH4 : 47a6b : 5 dư 3
\(\Rightarrow b\in3;8\)
\(để47a63⋮9\Rightarrow4+7+a+6+3⋮9\)
\(\Rightarrow a=7\)
\(để47a68⋮9\Rightarrow4+7+a+6+8⋮9\)
\(\Rightarrow a=2\)
HỌC TỐT !

B2 :P Ta có : \(B=\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(C=\frac{2^{10}-1}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Nên : B > C

a) ta có \(\frac{5}{24};\frac{15}{24};\frac{5}{8}\)
=>\(\frac{5}{24}< \frac{15}{24}< \frac{20}{24}\)quy đồng lên
b)\(\frac{4}{9};\frac{6+9}{6\cdot9};\frac{2}{3}\)
=>\(\frac{4}{9};\frac{15}{54};\frac{2}{3}\)
=>\(\frac{24}{54};\frac{15}{54};\frac{36}{54}\)
=>\(\frac{15}{54}< \frac{24}{54}< \frac{36}{54}\)

a/ Ta có :
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Leftrightarrow5^{36}>11^{24}\)
b/ Ta có :
\(5^{23}< 6.5^{22}\)
a, \(5^{36}\) và \(11^{24}\)
Ta có:
\(5^{36}\) = \(\left(5^3\right)^{12}\)= \(125^{12}\)
Và \(11^{24}\) = \(\left(11^2\right)^{12}\)= \(121^{12}\)
vì \(125^{12}\)> \(121^{12}\)
Nên \(5^{36}\) > \(11^{24}\)
b, \(5^{23}\) và 6. \(5^{22}\)
Mà \(5^{23}\)= 5. \(5^{22}\)
=> 5. \(5^{22}\) < 6. \(5^{22}\)
Nên \(5^{23}\) < 6. \(5^{22}\)

a
4 =22
5 =5.1
6=2.3
\(\Rightarrow BCNN\left(4,5,6\right)=2^2.3.5=60\)
BC (4,5,6 ) = B (60) ={0 ;60;120,240,360,420,......}
x-1 = {1 :61;121:241;361;421 ;.......}
mà x <400
=> x = 361

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, Ta có : \(\frac{5+10}{24}=\frac{15}{24}\)
\(\frac{5}{8}=\frac{5.3}{8.3}=\frac{15}{24}\)
\(\Rightarrow\frac{5}{24}< \frac{5+10}{24}=\frac{5}{8}\)
b,Ta có : \(\frac{4}{9}=\frac{4.6}{9.6}=\frac{24}{54}\)
\(\frac{6+9}{6.9}=\frac{15}{54}\)
\(\frac{2}{3}=\frac{2.18}{3.18}=\frac{36}{54}\)
\(\Rightarrow\frac{15}{54}< \frac{24}{54}< \frac{36}{54}\)
\(\Rightarrow\frac{6+9}{6.9}< \frac{4}{9}< \frac{2}{3}\)
a) \(\frac{5+10}{24}=\frac{15}{24}\); \(\frac{5}{8}=\frac{5.3}{8.3}=\frac{15}{24}\)
\(\Rightarrow\frac{5}{24}< \frac{15}{24};\frac{5+10}{24}=\frac{5}{8}\)
b) \(\frac{6+9}{6.9}=\frac{15}{54}\); \(\frac{4}{9}=\frac{24}{54};\frac{2}{3}=\frac{36}{54}\)
\(\Rightarrow\frac{15}{54}< \frac{24}{54}< \frac{36}{54}\)\(\Rightarrow\frac{6+9}{6.9}< \frac{4}{9}< \frac{2}{3}\)