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i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=\)\(2345-1000\div\left[19-2.3^2\right]\)
\(=\)\(2345-1000\div\left[19-2.9\right]\)
\(=\)\(2345-1000\div\left[19-18\right]\)
\(=\)\(2345-1000\div1\)
\(=\)\(2345-1000\)
\(=\)\(1345\)
j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=\)\(128-\left[68+8.2^2\right]\div4\)
\(=\)\(128-\left[68+8.4\right]\div4\)
\(=\)\(128-\left[68+32\right]\div4\)
\(=\)\(128-100\div4\)
\(=\)\(128-25\)
\(=\)\(3\)
k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+67\right\}\div15\)
\(=\)\(107-105\div15\)
\(=\)\(107-7\)
\(=\)\(7\)
b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)
\(=\)\(307-\left[20\div4+9\right]\div2\)
\(=\)\(307-\left[5+9\right]\div2\)
\(=\)\(307-14\div2\)
\(=\)\(307-7\)
\(=\)\(300\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
\( [(8𝑥 - 12) ∶ 4]. 33 = 36\)
\([(8𝑥-12)∶4]=36:33\)
\([(8𝑥-12)∶4]=\frac{36}{33}\)
\([(8𝑥-12)∶4]=\frac{12}{11}\)
\((8𝑥-12)=\frac{12}{11}.4\)
\((8𝑥-12)=\frac{48}{11}\)
\(8x=\frac{48}{11}+12\)
\(8x=\frac{48}{11}+\frac{132}{11}\)
\(8x=\frac{180}{11}\)
\(x=\frac{180}{11}\text{:8}\)
\(x=\frac{180}{11}\frac{.1}{8}\)
\(x=\frac{180}{88}\)
\(x=\frac{45}{22}\)
Vậy \(x=\frac{45}{22}\)
[(8x−12)∶4].33=36
\(\left[\right. \left(\right. 8 x - 12 \left.\right) ∶ 4 \left]\right. = 36 : 33\)
\(\left[\right. \left(\right. 8 x - 12 \left.\right) ∶ 4 \left]\right. = \frac{36}{33}\)
\(\left[\right. \left(\right. 8 x - 12 \left.\right) ∶ 4 \left]\right. = \frac{12}{11}\)
\(\left(\right. 8 x - 12 \left.\right) = \frac{12}{11} . 4\)
\(\left(\right. 8 x - 12 \left.\right) = \frac{48}{11}\)
\(8 x = \frac{48}{11} + 12\)
\(8 x = \frac{48}{11} + \frac{132}{11}\)
\(8 x = \frac{180}{11}\)
\(x = \frac{180}{11} :\text{8}\)
\(x = \frac{180}{11} \frac{. 1}{8}\)
\(x = \frac{180}{88}\)
\(x = \frac{45}{22}\)
Vậy \(x = \frac{45}{22}\)
a,5.(x-4) mũ 2-7=13
5.(x-4) mũ 2 =13+7
5.(x-4) mũ 2 =20
(x-4) mũ 2 = 20:5
(x-4)mũ 2= 4
(x-4) mũ 2=2 mũ 2
x-4=2
x=6
Vậy x = 6
b, phần này hình như thiếu gì đó
c,2 mũ x+ 3 - 3.2 mũ x+ 1=32
2 mũ x . 2 mũ 3 - 3 . 2 mũ x . 2 = 32
2 mũ x . 8 -( 3.2).2 mũ x = 32
2 mũ x . 8 -6 . 2 mũ x =32
2 mũ x .( 8- 6) = 32
2 mũ x = 32 : 2
2 mũ x = 16
2 mũ x=2 mũ 4
x = 4
vậy x = 4
k cho mình nha !!!
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
Bài 7:
a; 3\(^{x}\).3 = 243
3\(^{x+1}\) = 3\(^5\)
\(x+1\) = 5
\(x=5-1\)
\(x=4\)
Vậy \(x=4\)
b; 2\(^{x}\).162 = 1024
\(2^{x}\) = 1024 : 162
2\(^{x}\) = \(\frac{512}{81}\) (loại vì 512/81 không phải là số tự nhiên)
Không có số tự nhiên nào của x thỏa mãn đề bài.
Vậy \(x\in\) ∅
c; 64.4\(^{x}\) = 168
4\(^{x}\) = 168 : 64
4\(^{x}\) = \(\frac{21}{8}\) (loại)
vì 21/8 không phải là số tự nhiên
Vậy \(x\in\) ∅
d; \(2^{x}\) = 16
\(2^{x}\) = 2\(^4\)
\(x=4\)
Vậy \(x=4\)
Bài 8:
a; (\(2^{17}\) + 17\(^2\)).(9\(^{15}\) - 3\(^{15}\)).(2\(^4\) - 4\(^2\))
= (\(2^{17}\) + 17\(^2\)).(9\(^{15}\) - 3\(^{15}\)).(16 - 16)
= (\(2^{17}\) + 17\(^2\)).(9\(^{15}\) - 3\(^{15}\)).0
= 0
b; (\(8^{2017}\) - 8\(^{2015}\)):(8\(^{2014}\).8\(\))
= (\(8^{2017}\) - 8\(^{2015}\)): \(8^{2015}\)
= \(8^{2017}:8^{2015}\) - \(8^{2015}\) : 8\(^{2015}\)
= 8\(^2\) - 1
= 64 - 1
= 63
c; (1\(^3\) + 2\(^3\) + 3\(^4\) + 4\(^5\)).(\(3^8-81^2\))
= (1\(^3\) + 2\(^3\) + 3\(^4\) + 4\(^5\)).(3\(^8\) - 3\(^8\))
= (1\(^3\) + 2\(^3\) + 3\(^4\) + 4\(^5\)).0
= 0
d; (2\(^8\) + 8\(^3\)) : (2\(^5\).2\(^3\))
= (2\(^8\) + 2\(^9\)):(2\(^8\))
= 2\(^8\) : 2\(^8\) + 2\(^9\) : 2\(^8\)
= 1 + 2
= 3
Ai làm đúng mình tick cho ạ
Bài 9:
a; \(125^5\) : 25\(^3\)
= 5\(^{15}\) : 5\(^6\)
= 5\(^9\)
b; 27\(^6\) : 9\(^3\)
= 3\(^{18}\) : 3\(^6\)
= 3\(^{18}\) : 3\(^6\)
= 3\(^{12}\)
c; 4\(^{20}\) : 2\(^{15}\)
= 2\(^{40}\) : 2\(^{15}\)
= 2\(^{25}\) \(\)
câu 9:
d; 24\(^{n}\) : 2\(^{2n}\)
= 2\(^{3n}\) . 3\(^{n}\) : 2\(^{2n}\)
= (2\(^{3n}\) : 2\(^{2n}\)).3\(^{n}\)
= 2\(^{n}\).3\(^{n}\)
= 6\(^{n}\)
e; \(64^4\).16\(^5\).4\(^{20}\)
= 2\(^{12}\).2\(^{20}\).2\(^{40}\)
= 2\(^{72}\)
g; 32\(^4\) : 8\(^6\)
= 2\(^{20}\) : 2\(^{18}\)
= 2\(^2\)
= 4
Bài 10:
a; 2\(^{x}\).4 = 128
2\(^{x}\) = 128 : 4
2\(^{x}\) = 32
2\(^{x}\) = 2\(^5\)
\(x\) = 5
b; (2\(x\) + 1)\(^3\) = 125
(2\(x\) + 1)\(^3\) = 5\(^3\)
2\(x+1\) = 5
2\(x\) = 5 - 1
2\(x\) = 6
\(x\) = 6 : 2
\(x\) = 3
Vậy \(x=3\)
c; 2\(x\) - 2\(^6\) = 3
2\(x\) - 64 = 6
2\(x\) = 6 + 64
2\(x\) = 70
\(x=70:2\)
\(x\) = 34
Vậy \(x\) = \(34\)
Bài 10:
d; 64.4\(^{x}\) = 45
4\(^{x}\) = 45 : 64
4\(^{x}\) = \(\frac{45}{64}\) (loại)
Vậy \(x\in\) ∅
e; 27.3\(^{x}\) = 243
3\(^{x}\) = 243 : 27
3\(^{x}\) = 9
3\(^{x}\) = 3\(^2\)
\(x=2\)
Vậy \(x=2\)
Bài 10
n; 3\(^{x}\) + 25 = 26.2\(^2\) + 2.3\(^0\)
3\(^{x}\) + 25 = 26.4 + 2.1
3\(^{x}\) + 25 = 104 + 2
3\(^{x}\) + 25 = 106
3\(^{x}\) = 106 - 25
\(3^{x}\) = 81
3\(^{x}\) = 3\(^4\)
\(x=4\)
Vậy \(x=4\)
g; 49.7\(^{x}\) = 2401
7\(^{x}\) = 2401 : 49
7\(^{x}\) = 49
7\(^{x}\) = 7\(^2\)
\(x=2\)
Vậy \(x=2\)
Câu 10:
h; 3\(^{x}\) = 81
\(3^{x}\) = 3\(^4\)
\(x=4\)
Vậy \(x=4\)
k; 3\(^4\).3\(^{x}\) = 3\(^7\)
3\(^{x}\) = 3\(^7\) : 3\(^4\)
\(x=\) \(3^3\)
\(x=27\)
Vậy \(x=27\)
Bài 11:
a; 2\(^6\) và 8\(^3\)
2\(^6\) = (2\(^3\))\(^2\) = 8\(^3\)
Vậy \(2^6\) = 8\(^3\)
5\(^3\) và 3\(^5\)
5\(^3\) = 125; 3\(^5\) = 243
125 < 243
5\(^3\) < \(3^5\)
3\(^2\) và 2\(^3\)
3\(^2\) = 9
2\(^3\) = 8
3\(^2\) > 2\(^3\)
2\(^6\) và 6\(^2\)
2\(^6\) = 64; 6\(^2\) = 36
64 > 36
2\(^6\) > 6\(^2\)
Bài 11:
b; A = 2009.2011 và B = 20102
20102 < 20110 = 2011 x 10 < 2011 x 2009
Vậy 20102 < 2009 x 2011
A > B
c; A = 2015 x 2017 và B = 2016 x 2016
A = 2015 x 2017
A = (2016 - 1) x (2016 + 1)
A = 2016 x 2016 + 2016 - 2016 - 1
A = 2016 x 2016 + (2016 - 2016) - 1
A = 2016 x 2016 + 0 - 1
A = 2016 x 2016 - 1 < 2016 x 2016
A < B
a; 3\(^{x}\).3 = 243
3\(^{x + 1}\) = 3\(^{5}\)
\(x + 1\) = 5
\(x = 5 - 1\)
\(x = 4\)
Vậy \(x = 4\)
b; 2\(^{x}\).162 = 1024
\(2^{x}\) = 1024 : 162
2\(^{x}\) = \(\frac{512}{81}\) (loại vì 512/81 không phải là số tự nhiên)
Không có số tự nhiên nào của x thỏa mãn đề bài.
Vậy \(x \in\) ∅
c; 64.4\(^{x}\) = 168
4\(^{x}\) = 168 : 64
4\(^{x}\) = \(\frac{21}{8}\) (loại)
vì 21/8 không phải là số tự nhiên
Vậy \(x \in\) ∅
d; \(2^{x}\) = 16
\(2^{x}\) = 2\(^{4}\)
\(x = 4\)
Vậy \(x = 4\)
của bạn đây
Bài 11 câu d:
A = 2017\(^0\) và B = 1\(^{2017}\)
A = 2017\(^0\) = 1 và B = 1\(^{2017}\) = 1
A = B
Bài 12:
A = 1 + 2 + 2\(^2\) + .... + 2\(^{2007}\)
A x 2 = 2 + 2\(^2\) +...+2\(^{2007}\) + 2\(^{2008}\)
A x 2 = (2 + 2\(^2\)+...+ 2\(^{2007}\)+ 2\(^{2008}\))-(1+2+...+2\(^{2007}\))
A x 2 - A = 2+2\(^2\)+...+2\(^{2007}\)+2\(^{2018}\)- 1 - 2 -...-\(2^{2007}\)
A x 2 - A = (2-2)+(2\(^2\)-2\(^2\))+...+(2\(^{2007}\)-2\(^{2007}\))+(2\(^{2008}\) -1)
A = 0+ 0+ 0+...+ 0 + (2\(^{2008}\) - 1)
A = 2\(^{2008}\) - 1
Bài 13:
A = 1+ 3 + 3\(^2\) + ... + 3\(^6\) + 3\(^7\)
A x 3 = 3 x (1+ 3+ 3\(^2\) + ...+ 3\(^7\))
A x 3 = 3 + 3\(^2\) + 3\(^3\) + ...+ 3\(^8\)
A x 3 - A = 3+3\(^2+3^3+..+3^8\) -(1+ 3 + 3\(^2\)+...+3\(^7\))
2A = 3\(^2\)+ 3\(^3\) + ... + 3\(^8\)-1 - 3- 3\(^2\) - ...- 3\(^7\)
2A = (3\(^2\) - 3\(^2\)) + (3\(^3\)-3\(^3\)) +... +(3\(^7\) - 3\(^7\))+(3\(^8-1\))
2A = 0 + 0 + ... + 0 + 3\(^8\) - 1
2A = 3\(^8\) - 1
A = (3\(^8\) - 1) : 2
Bài 14:
A = 1 + 3 + 3\(^2\) + ... + 3\(^{2006}\)
3 x A = 3 + 3\(^2\) + ... + 3\(^{2006}\) + 3\(^{2007}\)
3A - A = 3 + 3\(^2\) + ... + 3\(^{2006}\) + 3\(^{2007}\) -(1 + 3 + 3\(^2\) + ... + 3\(^{2006}\))
2A = 3 + 3\(^2\) + ... + 3\(^{2007}\)-1-3-3\(^2\) -...-\(3^{2006}\)
2A = (3-3)+(3\(^2-3^2\))+...+(3\(^{2006}\)-3\(^{2006}\))+(3\(^{2007}\) -1)
2A = 0+0+...+0+3\(^{2007}\) - 1
A = (3\(^{2007}\) - 1):2
Bài 19:
a; 5\(^{36}\) và 11\(^{24}\)
(5\(^3\))\(^{12}\) = 125\(^{12}\)
11\(^{24}\) = (11\(^2\))\(^{24}\) = 121\(^{12}\)
125 > 121 nên
125\(^{12}\) > 121\(^{12}\)
b; 3\(^{2n}\) và 2\(^{3n}\)
3\(^{2n}\) = (3\(^2\))\(^{n}\) = 9\(^{n}\) > 8\(^{n}\) = (2\(^3\))\(^{n}\) = 8\(^{n}\)
Vậy 3\(^{2n}\) > 2\(^{3n}\)