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A= 150-(100-99+98-97+...-3+2-1)
từ 1-100 có 100 SH. Ta nhóm 4 số vs nhau như sau : (100-00+98-87)+(...)+(4-3+2-1)
Có tất cả số nhóm là : 100:4=25 nhóm. Mà mỗi nhóm ta tính có kết quả là 2, vậy tao có
A=150-(2.25)
A=150-50
A=100
a)=\(\frac{7}{30}\)-\(\frac{17}{12}\)=\(\frac{-71}{60}\) b) = \(\frac{29}{20}\)+\(\frac{-31}{30}\)+\(\frac{-2}{3}=\frac{-1}{4}\)
\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5-7.2\right)}\)\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-14\right)}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}=\frac{2.9}{-9}=-2\)
Vậy A=-2
a) \(3^{7} \cdot 27^{5} \cdot 81^{3}\)
- Đổi về cơ số 3: \(27 = 3^{3} , \textrm{ }\textrm{ } 81 = 3^{4}\).
\(3^{7} \cdot \left(\right. 3^{3} \left.\right)^{5} \cdot \left(\right. 3^{4} \left.\right)^{3} = 3^{7} \cdot 3^{15} \cdot 3^{12} = 3^{7 + 15 + 12} = 3^{34} .\)
b) \(100^{6} \cdot 1000^{5} \cdot 10 \textrm{ } 000^{3}\)
- Đổi về cơ số 10: \(100 = 10^{2} , \textrm{ }\textrm{ } 1000 = 10^{3} , \textrm{ }\textrm{ } 10 \textrm{ } 000 = 10^{4}\).
\(\left(\right. 10^{2} \left.\right)^{6} \cdot \left(\right. 10^{3} \left.\right)^{5} \cdot \left(\right. 10^{4} \left.\right)^{3} = 10^{12} \cdot 10^{15} \cdot 10^{12} = 10^{39} .\)
c) \(\frac{36^{5}}{18^{5}}\)
- Do cùng số mũ:
\(\left(\left(\right. \frac{36}{18} \left.\right)\right)^{5} = 2^{5} = 32.\)
d) \(24 \cdot 5^{2} + 5^{2} \cdot 5^{3}\)
\(= 24 \cdot 25 + 25 \cdot 125 = 600 + 3125 = 3725.\)
e) \(\frac{125^{4}}{5^{8}}\)
- Đổi về cơ số 5: \(125 = 5^{3}\).
\(\left(\right. 5^{3} \left.\right)^{4} : 5^{8} = 5^{12} : 5^{8} = 5^{12 - 8} = 5^{4} = 625.\)
TÓM lại là a) \(3^{34}\)
b) \(10^{39}\)
c) \(32\)
d) \(3725\)
e) \(625\)
bạn muốn chép đáp án hay sem cách làm///??
a: \(3^7\cdot27^5\cdot81^3\)
\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)
\(=3^7\cdot3^{15}\cdot3^{12}=3^{7+15+12}=3^{34}\)
b: \(100^6\cdot1000^5\cdot10000^3\)
\(=\left(10^2\right)^6\cdot\left(10^3\right)^5\cdot\left(10^4\right)^3\)
\(=10^{12}\cdot10^{15}\cdot10^{12}=10^{15+12+12}=10^{39}\)
c: \(36^5:18^5=\left(\frac{36}{18}\right)^5=2^5=32\)
d: \(24\cdot5^2+5^2\cdot5^3\)
\(=5^2\left(24+5^3\right)\)
\(=25\cdot\left(24+125\right)=25\cdot149=3725\)
e: \(125^4:5^8=\left(5^3\right)^4:5^8=5^{12}:5^8=5^{12-8}=5^4\)
a: \(=36:4+2\cdot25=9+50=59\)
b: \(=79\left(82+18\right)=79\cdot100=7900\)
c: \(=49-9-\left(4^2+2^2\right)\)
\(=40-16-4=40-20=20\)
d: \(=16+\left[400:\left(200-42-138\right)\right]\)
\(=16+400:20=16+20=36\)
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
a, 600 : { 450 : 450 - 4 . 5 3 - 2 3 . 5 2 }
= 600:{450:[450 - (4.125 - 8.25)]} = 600:{450:[450 - (500 - 200)]}
= 600:{450:[450 - 300]} = 600:{450:150} = 600:3 = 200
b, 18 . { 420 : 6 - 150 - ( 68 . 2 - 2 3 . 5 ) }
= 18.{70 - [150 - (68.2 - 8.5)]} = 18.{70 - [150 - (136 - 40)]}
= 18.{70 - [150 - 96]} = 18.{70 - 54} = 18.16 = 288