Bài 1: Tính ( Nhanh nếu có thể )
a) \(25.8^3-23.8^3\)
b) \(5^4-2.5^3\)
c) \(2.4^3-4^3.7-6.4^3\)
d) \(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right]\)
Bài 1: Tính. ( Nhanh nếu có thể )
a) \(25.8^3-23.8^3\)
b) \(5^4-2.5^3\)
c) \(2.4^3-4^3.7-6.4^3\)
d) \(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right].10^3\)
a) \(25.8^3-23.8^3=8^3\left(25-23\right)\)
\(=8^3.2\)
\(=2^9.2=2^{10}\)
b) \(5^4-2.5^3=5^3.5-2.5^3\)
\(=5^3\left(5-2\right)\)
\(=5^3.3=375\)
c)\(2.4^3-4^3.7-6.4^3=4^3\left(2-7-6\right)\)
\(=4^3.-11=-704\)
d)\(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right].10^3\)
\(=3^2.10^3-\left[13^2-2^2\left(5^2+15\right)\right].10^3\)
\(=3^2.10^3-\left[13^2-2^2.40\right].10^3\)
\(=10^3\left[3^2-9\right]\)
\(=0\)
A=\(^{3^2.10^2-\left[13^2-\left(5^2.4+2^2.15\right)\right].10^3}\)
\(^{3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right].10^3}\)
Thực hiện các phép tính (tính nhanh nhất nếu có thể):
\(A,\)\(2^2.5-\left(1^{10}+8\right):3^2\)
\(B, 5^8:5^6+4.(3^2-1)\)
\(C,\)\(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)
a: \(2^2\cdot5-\dfrac{\left(1^{10}+8\right)}{3^2}\)
\(=4\cdot5-\dfrac{1+8}{3}\)
=20-3
=17
b: \(5^8:5^6+4\left(3^2-1\right)\)
\(=5^2+4\left(9-1\right)\)
=25+4*8
=25+32
=57
c: \(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)
\(=400-36+20:\left[27-5\right]\)
\(=364+\dfrac{20}{22}\)
\(=364+\dfrac{10}{11}=\dfrac{4014}{11}\)
A) 2².5 - (1¹⁰ + 8) : 3²
= 4.5 - (1 + 8) : 9
= 20 - 9 : 9
= 20 - 1
= 19
B) 5⁸ : 5⁶ + 4.(3² - 1)
= 5² + 4.(9 - 1)
= 25 + 4.8
= 25 + 32
= 57
C) 400 - {36 - 20 : [3³ - (8 - 3)]}
= 400 - [36 - 20 : (27 - 5)]
= 400 -(36 - 20 : 22)
= 400 - (36 - 10/11)
= 400 - 386/11
= 4014/11
Bài 1: Tính:
\(a,\left(0,25\right)^3.32\) \(b,\left(0,125\right)^3.512\) \(c,\dfrac{8^2.4^5}{2^{20}}\) \(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}\)
Bài 2: Tìm giá trị nhỏ nhất của các biểu thức sau:
\(a,A=\left|x-\dfrac{3}{4}\right|\) \(b,B=1,5+\left|2-x\right|\) \(c,A=\left|2x-\dfrac{1}{3}\right|+107\) \(d,M=5\left|1-4x\right|-1\)
Bài 3: Tìm giá trị lớn nhất của biểu thức sau:
\(a,C=-\left|x-2\right|\) \(b,D=1-\left|2x-3\right|\) \(c,D=-\left|x+\dfrac{5}{2}\right|\)
(mn giải giúp mk với, thanks mn nhìu!)
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
bài 2 : Tính giá trị biểu thức sau :
a) \(\frac{6^2.6^3}{3^5}\) b) \(\frac{25^2.4^2}{5^5.\left(-2\right)^5}\) c) \(\frac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}\) d) \(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)
Mình cần rất gấp ạ nên bn nào nhanh mik tick nhé !
a,32
b,\(-\frac{1}{10}\)
c,-1000000
d,\(\frac{9}{16}\)
\(3.8.5^2+2.4^3.12+\left(2^3+3\right).6.4\)
\(600:\left\{450-\left[450-\left(2^3.5^2\right)\right]\right\}\)
Bài 1: So sánh:
a) 5^255 và 2^512 ( 2 cách)
b) 8^12 và 12^8
Bài 2: Chứng minh rằng:
a) A = \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}<1\)
b) B = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
c) C = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}<\frac{3}{4}\)
Tính giá trị của các biểu thức sau:
a)\(\frac{6^2.6^3}{3^5}\) b)\(\frac{25^2.4^2}{5^5.\left(-2\right)^5}\) c)\(\frac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}\) d)\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)