Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) \(2^8.2.4=2^9.2^2=2^{11}\)
b) \(8^5:64=8^5:8^2=8^3\)
c) \(3^7:9=3^7:3^2=3^5\)
d) \(9^{17}.81=9^{17}.9^2=9^{19}\)
e) \(x^6.x.x^2=x^9\)
Bài 2:
a) \(2^x-15=17\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5
b) \(2.3^x=162\)
\(3^x=162:2\)
\(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy x = 4
c) \(5.x.5^2=10\)
\(\Rightarrow x.5^3=10\)
\(\Rightarrow x.125=10\)
\(\Rightarrow x=10:125\)
\(\Rightarrow x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\)
d) \(5.x^2-1=124\)
\(\Rightarrow5.x^2=125\)
\(\Rightarrow x^2=125:5\)
\(\Rightarrow x^2=5^2\)
\(\Rightarrow x=\pm5\)
Vậy \(x=\pm5\)
Câu 1:
a)28.2.4=28.2.22=211
b)85:64=85:82=83
c)37:9=37:32=35
d)917.81=917.92=919
e)x6.x.x2=x9
5891 :\(\left\{928-\left(247-82\right).5\right\}\)
=5891:\(\left\{928-165.5\right\}\)
=5891:\(\left\{928-825\right\}\)
=5891:103
=57.194175
a, Ta có: \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^4}\right)^7=\left(\dfrac{1}{3}\right)^{28}=\dfrac{1}{3^{28}}\)
\(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3^5}\right)^6=\left(\dfrac{1}{3}\right)^{30}=\dfrac{1}{3^{30}}\)
Vì \(\dfrac{1}{3^{28}}>\dfrac{!}{3^{30}}\Rightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\Rightarrow\) \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
b, Ta có: \(\left(\dfrac{3}{8}\right)^5=\dfrac{3^5}{\left(2^3\right)^5}=\dfrac{243}{2^{15}}>\dfrac{243}{3^{15}}>\dfrac{125}{3^{15}}=\dfrac{5^3}{\left(3^5\right)^3}=\left(\dfrac{5}{243}\right)^3\)
\(\Rightarrow\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
nên các bn giúp mk nha,còn nhìu ý nữa cơ 
mai hoc rùi
Bài 1: Tính:
a) 27 : 22 + 54 : 53. 24 - 3. 25
= 25 + 5 . 24 - 3 . 25
= 32 + 5 . 16 - 3 . 32
= 32 + 80 - 96
= 112 - 96
= 16
b) ( 37 . 35) : 310+ 5 . 24 - 73 : 7
= 312 : 310 + 5 . 24 - 72
= 32 + 5 . 24 - 72
= 9 + 5 . 16 - 49
= 9 + 80 - 49
= 89 - 49
= 40
Bài 2: Tính hợp lí:
a) ( 62007 - 62006 ) : 62006
= 62007 : 62006 - 62006 : 62006
= 6 - 1
= 5
b) ( 112003 + 112002 ) : 112002
= 11 + 1
= 12
c) 320 : ( x3 - 24 ) + 24 = 32
320 : ( x3 - 24 ) = 32 - 24 = 8
x3 - 24 = 320 : 8
x3 - 24 = 40 + 24
x3 = 64
x3 = 43 = 4
d) 130 - ( 100 + x ) = 25
( 100 + x ) = 103 - 25
100 + x = 105 - 100
x = 5
Bn ơi đừng tự ti như vậy nha !!! Mỗi người đều có một khuyết điểm mà, tri thức luôn rộng lớn bao la. Hãy làm việc đó bằng cách bn tự làm những bài kia nha.
Chúc bn hc tốt môn toán :))
2)
a) \(\left(6^{2007}-6^{2006}\right):6^{2006}\)
\(=\left(6^{2006}.6-6^{2006}.1\right):6^{2006}\)
\(=\left[6^{2006}.\left(6-1\right)\right]:6^{2006}\)
\(=6^{2006}:6^{2006}.5\)
\(=5\)
b) \(\left(11^{2003}+11^{2002}\right):11^{2002}\)
\(=\left(11^{2002}.11+11^{2002}.1\right):11^{2002}\)
\(=\left[11^{2002}.\left(11+1\right)\right]:11^{2002}\)
\(=11^{2002}:11^{2002}.12\)
\(=12\)
c) \(130:\left(x^3-24\right)+24=32\)
\(\Leftrightarrow130:\left(x^3-24\right)=32-24\)
\(\Leftrightarrow130:\left(x^3-24\right)=8\)
\(\Leftrightarrow x^3-24=\dfrac{65}{4}\)
\(\Leftrightarrow x^3=\dfrac{65}{4}+24\)
\(\Leftrightarrow x^3=\dfrac{161}{4}\)
\(\Leftrightarrow x=\sqrt[3]{\dfrac{161}{4}}\)
Vậy \(x=\sqrt[3]{\dfrac{161}{4}}\)
d) \(130-\left(100+x\right)=25\)
\(\Leftrightarrow100+x=130-25\)
\(\Leftrightarrow100+x=105\)
\(\Leftrightarrow x=105-100\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
a) 516 : 514 + 24 . 2 - 20100
= 52 + 16 . 2 - 1
= 25 + 32 - 1
= 57 - 1
= 56
b) 62 - 22 . 3 + 16 . 3
= 36 - 4 . 3 + 48
= 32 . 3 + 48
= 96 + 48
= 144
a.
\(\frac{6}{7}+\frac{5}{8}\div5-\frac{3}{16}\times\left(-2\right)^2=\frac{6}{7}+\frac{5}{8}\times\frac{1}{5}-\frac{3}{16}\times4=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{48}{56}+\frac{7}{56}-\frac{42}{56}=\frac{13}{56}\)
b.
\(\frac{2}{3}+\frac{1}{3}\times\left(-\frac{4}{9}+\frac{5}{6}\right)\div\frac{7}{12}=\frac{2}{3}+\frac{1}{3}\times\left(-\frac{8}{18}+\frac{15}{18}\right)\times\frac{12}{7}=\frac{2}{3}+\frac{7}{18}\times\frac{4}{7}=\frac{2}{3}+\frac{2}{9}=\frac{6}{9}+\frac{2}{9}=\frac{8}{9}\)
c.
\(\left[6+\left(\frac{1}{3}\right)^3-\left|-2\right|\right]\div\frac{3}{2}=\left(6+\frac{1^3}{3^3}-2\right)\times\frac{2}{3}=\left(4+\frac{1}{27}\right)\times\frac{2}{3}=\left(\frac{108}{27}+\frac{1}{27}\right)\times\frac{2}{3}=\frac{109}{27}\times\frac{2}{3}=\frac{218}{81}\)
d.
\(\left(-2\right)^3\times\left(-\frac{1}{24}\right)+\left(\frac{4}{3}-\frac{11}{6}\right)\div\frac{5}{12}\)\(=\left(-8\right)\times\left(-\frac{1}{24}\right)+\left(\frac{8}{6}-\frac{11}{6}\right)\times\frac{12}{5}=\frac{1}{3}+\left(-\frac{3}{6}\right)\times\frac{12}{5}=\frac{1}{3}-\frac{6}{5}=\frac{5}{15}-\frac{18}{15}\)\(=-\frac{13}{15}\)