c) Ta có: x+11 chia hết cho x+1

x+1+10 chia hết cho x+1

Vì x+1 chia hết cho x+1 nên 10 chia hết cho x+1.

Vậy x+1\(\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)

\(\Rightarrow x\in\left\{0;1;4;9\right\}\)

\(2^3.3^2.28+2^3.3^5.8\\ =2^3.3^2.2^2.7+2^3.3^5.2^3\\ =2^5.3^2.7+2^6.3^5\\ =2^5.3^2\left(7+2.3^3\right)\\ =32.9\left(7+54\right)\\ =32.9.61=17568\)

\(2^3.3^2.28+2^3.3^5.28\)

=> \(28.(2^3.3^2+2^3.3^5)\)

=> \(28.[3^3(3^2+3^5)]\)

=> \(28.[3^3.252]\)

=> ............( tự tính)

a) 15+25+35+...+115

(115-15):5+1=21 số hạng

(115+15).21:2=1365

a: Số số hạng là (115-15):10+1=11(số)

Tổng là (115+15)*11/2=65*11=715

b: =31(64+146)+210

=210(31+1)

=210*32=6720

sai đề 100%

ko có quy luật

biết rồi mi

\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+....+100\left(101-1\right)\)

\(=1.2+2.3+3.4+...+100.101-\left(1+2+3+...+100\right)\)

Ta có: \(M=1.2+2.3+...+100.101\)

\(3M=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+100.101\left(102-99\right)\)

\(=-0.1.2+1.2.3-1.2.3+2.3.4-...-99.100.101+100.101.102\)

\(=100.101.102\)

\(\Rightarrow M=\frac{100.101.102}{3}=343400\)

\(N=1+2+...+100=\frac{\left(100+1\right).100}{2}=5050\)

\(B=M-N=338350\)

a)271.11

\(a.271.11-17.11\)

\(=\left(271-17\right).11\)

\(=254.11\)

\(=2794\)

1/\(a,271\cdot11-17\cdot11=11\left(271-17\right)=11\cdot200=2200\)
\(b,136\cdot23+136\cdot17+40\cdot36=136\left(23+17\right)+40\cdot36=136\cdot40+40\cdot36=40\left(136+36\right)=40\cdot172=6880\)
2/\(a,17\cdot6^2-230:x=584\Rightarrow17\cdot36-230:x=584\Rightarrow612-230:x=584\Rightarrow230:x=612-584=28\Rightarrow x=230:28=\frac{230}{28}=\frac{115}{14}\)
\(b,293-2^3\cdot3^3+x:25=80\Rightarrow293-8\cdot27+x:25=80\Rightarrow293-216-80+x:25=0\Rightarrow x:25=3\Rightarrow x=3\cdot25=75\)

\(c,4\cdot5^2-x:2=9\Rightarrow4\cdot25-9=x:2\Rightarrow x:2=91\Rightarrow x=91\cdot2=182\)

a, \(4^7.3^4.9^6:6^{13}\)

\(=\left(2^{14}.3^4.3^{12}\right):\left(2^{13}.3^{13}\right)\)

\(=2^{14}:2^{13}.3^{16}:3^{13}\)

\(=2.3^3=54\)

b, \(2^3.3^2-5^{16}:25^7\)

\(=72-5^{16}:5^{14}\)

\(=72-5^2=47\)