Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(x=5\)
2e)Đặt \(A=1+3+3^2+...+3^{200}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{201}\)
\(\Rightarrow2A=3^{201}-1\)
\(\Rightarrow A=\frac{3^{201}-1}{2}\)
\(\Rightarrow A< 3^{201}\)
Hay \(1+3+3^2+...+3^{200}< 3^{201}\)
,bn nào giỏi toán giúp mk nha
a) \(3.5^2-16:2^3.2\)
\(=3.25-16:8.2\)
\(=75-2.2\)
\(=75-4\)
\(=71\)
b) \(168+\left\{\left[2\left(2^4+3^2\right)-256^0\right]:7^2\right\}\)
\(=168+\left\{\left[2\left(16+9\right)-256^0\right]:7^2\right\}\)
\(=168+\left[\left(2.25-256^0\right):7^2\right]\)
\(=168+\left[\left(50-1\right):7^2\right]\)
\(=168+\left(49:7^2\right)\)
\(=168+\left(49:49\right)\)
\(=168+1\)
\(=169\)
c) \(9^{20}:9^{18}-\left(4^2-7\right)^2+8.5^2+5600:\left(3^3+1^8\right)\)
\(=9^{20}:9^{18}-\left(16-7\right)^2+8.5^2+5600:\left(27+1\right)\)
\(=9^{20}:9^{18}-9^2+8.5^2+5600:28\)
\(=9^{20-18}-9^2+8.25+5600:28\)
\(=9^2-9^2+200+200\)
\(=81-81+200+200\)
\(=200+200\)
\(=400\)
a) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4.125-4.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
b) \(109.5^2-3^2.25\)
\(=109.25-9.25\)
\(=25\left(109-9\right)\)
\(=25.100\)
\(=2500\)
c) \(\left[5^2.6-20.\left(37-2^5\right)\right]:10-20\)
\(=\left[5^2.6-20.\left(37-32\right)\right]:10-20\)
\(=\left(5^2.6-20.5\right):10-20\)
\(=\left(25.6-20.5\right):10-20\)
\(=\left(150-100\right):10-20\)
\(=50:10-20\)
\(=5-20\)
\(=-15\)
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
a) \(3^4.3^3=3^{4+3}=3^7\)
b) \(5^2.5^7=5^{2+7}=5^9\)
c) \(7^5.7=7^5.7^1=7^{5+1}=7^6\)
d) \(x^7.x.x^4=x^7.x^1.x^4=x^{7+1+4}=x^{12}\)
a,34.33 =34+3 =37
b,52.57 = 52+7 =59
c,75.7 = 75+1 =76
d,x7.x.x4 =x7+1+4 =x12
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)