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\(3^3.5^3-20.\left\{300-\left[546-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)
= \(27.125-20.\left\{300-\left[546-8.50\right]\right\}\)
= \(3375-20.\left\{300-\left[546-400\right]\right\}\)
= \(3375-20.\left\{300-146\right\}\)
= \(3375-20.154=3375-3080=295\)
Đáp án hoàn toàn chính xác đấy.
Chúc bạn học tốt :))
33.53-20.{ 300-[ 546-23.(78:76+70)]}
=-98-20.{300- [546-8.50]}
=-98-20.{300-[546-400]}
=-98-20.{300-146}
=-98-20.154
=-3178
Chúc bạn học tốt Khanh Nhí
\(3^3\cdot5^3-20\cdot\left\{300-\left[546-2^3\cdot\left(7^8-7^6+7^0\right)\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300-\left[546-2^3\cdot5647153\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300-\left[546-45177224\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300--45176678\right\}\)
\(=3^3\cdot5^3-20\cdot45176978\)
\(=3375-903539560\)
\(=-903536185\)
\(626500:\left\{50^2:\left[178-4\cdot\left(35-21:3\right)\right]\right\}\)
\(=626500:\left\{50^2:\left[178-4\cdot\left(35-7\right)\right]\right\}\)
\(=626500:\left\{50^2:\left[178-4\cdot28\right]\right\}\)
\(=626500:\left\{50^2:\left[178-112\right]\right\}\)
\(=626500:\left\{50^2:66\right\}\)
\(=626500:\frac{1250}{33}\)
\(=\frac{82698}{5}\)
CHUC BAN HOC TOT >.<
1
3\(^5\).5\(^3\)-20.{300-[546-2\(^3\).(7\(^8\):7\(^6\)+7\(^0\))]}
= 243.125-20.{300-[546-8.(49+1)
= 243.125-20.[300-(546-8.50)
= 243.125-20.[300-(546-400)
= 243.125-20.(300-146)
= 243.125-20.154
= 30375-3080
= 27295
32 . 53 - 20{300 - [540 - 23 (78 : 76 + 70)]}
= 9 . 125 - 20{300 - [540 - 8(72 + 70)]}
= 9 . 125 - 20{300 - [540 - 8(49 + 1)]
= 9 . 125 - 20{300 - [540 - 8.50]
= 9 . 125 - 20{300 - [540 - 400]}
= 9 . 125 - 20{300 - 140}
= 9 . 125 - 20. 160
= 1125 - 3200 = -2075
\(=9\cdot125-20\cdot\left\{300-\left[540-8\cdot\left(7^2+1\right)\right]\right\}\)
\(=1125-20\cdot\left\{300-\left[540-8\cdot\left(49+1\right)\right]\right\}\)
\(=1125-20\cdot\left[300-\left(540-8\cdot50\right)\right]\)
\(=1125-20\cdot\left[300-\left(540-400\right)\right]\)
\(=1125-20\cdot\left(300-140\right)\)
\(=1125-20\cdot160\)
\(=1125-3200\)
\(=-2075\)
a)5.4.9-6^2=180-36=144
b)15:3-2^2=5-4=1
c)5.3^5:3^3-8.5=5.3^2-40=5.9-40=45-40=5
a)mk ko bt
b)15 * 35 / (35 / 34 ) - 23 *5
=15 / 3 - 4
=5 - 4
=1
c) 5 * 35 / (38 / 35 ) - 23 * 5
=5 * 35 / 27 - 23 * 5
=1215 / 27 - 40
=45 - 40
=5
d) 4* [(3 + 37 + 34)*10 + 97]-300
=4*[(3+2187+81)*10+97]-300
=4*[(2190+81)*10+97]-300
=4*[2271*10+97]-300
=4*[22710+97]-300
=4*22807-300
=9128-300
=90928
\(a)47-\left[\left(45.2^4-5^2.12\right):14\right]\)
\(=47-\left[\left(45.16-25.12\right):14\right]\)
\(=47-\left[\left(720-300\right):14\right]\)
\(=47-\left[420:14\right]\)
\(=47-30\)
\(=17\)
\(b)50-\left[\left(20-2^3\right):2+34\right]\)
\(=50-\left[\left(20-8\right)\right]:2+34\)
\(=50-\left[12:2+34\right]\)
\(=50-\left[6+34\right]\)
\(=50-40\)
\(=10\)
\(c)10^2-\left[60:\left(5^6:5^4-3,5\right)\right]\)
\(=10^2-\left[60:\left(5^2-3,5\right)\right]\)
\(=10^2-\left[60:\left(25-3,5\right)\right]\)
\(=10^2-\left[60:21,5\right]\)
\(=100-\dfrac{120}{43}\)
\(=\dfrac{4180}{43}\)
\(d)50-\left[\left(50-2^3.5\right):2+3\right]\)
\(=50-\left[\left(50-8.5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left[10:2+3\right]\)
\(=50-5+3\)
\(=50-8\)
\(=42\)
\(e)10-\left[\left(8^2-48\right).5+\left(2^3.10+8\right)\right]:28\)
\(=10-\left[\left(64-48\right).5+\left(8.10+8\right)\right]:28\)
\(=10-\left[16.5+\left(80+8\right)\right]:28\)
\(=10-\left[80+88\right]:28\)
\(=10-168:28\)
\(=10-6\)
\(=4\)
\(f)8697-\left[3^7:3^5+2.\left(13-3\right)\right]\)
\(=8697-\left[3^2+2.\left(13-3\right)\right]\)
\(=8697-\left[9+2.10\right]\)
\(=8697-9+20\)
\(=8697-29\)
\(=8668\)
\(g)2011+5\left[300-\left(17-7\right)^2\right]\)
\(=2011+5.\left[300-10^2\right]\)
\(=2011+5.\left[300-100\right]\)
\(=2011+5.200\)
\(=2011+1000\)
\(=3011\)
\(h)695-\left[200+\left(11-1\right)^2\right]\)
\(=695-\left[200+10^2\right]\)
\(=695-200+100\)
\(=695-300\)
\(=395\)
\(i)129-5\left[29-\left(6-1\right)^2\right]\)
\(=129-5\left[29-5^2\right]\)
\(=129-5\left[29-25\right]\)
\(=129-5.4\)
\(=129-20\)
\(=109\)
A=(5+5^2)+(5^3+5^4)+...(5^299+5^300)
A=5(1+5)+5^2(1+5)+...+5^299(1+5)
A=5.6+5^2.6+...+5^299.6 => Achia hết cho 6.
Tường tự phần A nhóm 3 số với nhau chia hết cho 31
phần B đường nhiên sẽ chia hết cho 7 vì mỗi số hạng đều chia hết cho 7, nhóm 2 số với nhau chia hết cho 8
\(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
\(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{5^4\cdot4^4\cdot5^4}{5^5\cdot5^5.4^5}=\frac{1}{100}\)
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2=3-1+\frac{1}{8}=\frac{17}{8}\)
câu này giống câu dưới mà Khanh Nhí
33.53-20.{300-[546-23.(78:76+70)]}
= 33.53-20.{300-[546-23.(72+1)]}
= 33.53-20.[300-(546-23.50)]
= 33.53-20.[300-(546-400)]
= 33.53-20.(300-146)
= 33.53-20.154
= 3375-3080
= 295.