a)

\(175\cdot19+38\cdot175+43\cdot175\\ =175\cdot19+175\cdot38+175\cdot43\\ =175\cdot\left(19+38+43\right)\\ =175\cdot100\\ =17500\)

b)

\(125\cdot75+125\cdot13-80\cdot125\\ =125\cdot75+125\cdot13-125\cdot80\\ =125\cdot\left(75+13-80\right)\\ =125\cdot10\\ =125\cdot8\\ =1000\)

a, 175. 19 + 38. 175 + 43. 175

= 175. 19 + 175. 38 + 175. 43

= 175.(19 + 38 + 43)

= 175. 100

= 17500 

Bài 5:

a: \(37\cdot146+46\cdot2-46\cdot37\)

\(=37\left(146-46\right)+46\cdot2\)

\(=37\cdot100+92=3700+92=3792\)

b: \(2\cdot5\cdot71+5\cdot18\cdot2+10\cdot11\)

\(=10\cdot71+10\cdot18+10\cdot11\)

\(=10\left(71+18+11\right)=10\cdot100=1000\)

c: \(135+360+65+40\)

=135+65+360+40

=200+400

=600

d: \(27\cdot75+25\cdot27-450\)

\(=27\left(75+25\right)-450\)

=2700-450

=2250

Bài 4:

a: \(32\cdot163+32\cdot837\)

\(=32\cdot\left(163+837\right)\)

\(=32\cdot1000=32000\)

b: \(2\cdot3\cdot4\cdot5\cdot25=2\cdot5\cdot4\cdot25\cdot3=3\cdot10\cdot100=3000\)

c: \(25\cdot27\cdot4=27\cdot100=2700\)

Bài 3:

a: \(128\cdot19+128\cdot41+128\cdot40\)

\(=128\cdot\left(19+41+40\right)=128\cdot100=12800\)

b: \(375+693+625+307\)

=375+625+693+307

=1000+1000

=2000

c: \(37+42-37+22\)

=37-37+42+22

=0+64

=64

d: \(21\cdot32+21\cdot68\)

\(=21\cdot\left(32+68\right)=21\cdot100=2100\)

Bài 2:

a: \(17\cdot85+15\cdot17-120\)

\(=17\left(85+15\right)-120\)

=1700-120

=1580

b: \(189+73+211+127\)

=189+211+73+127

=400+200

=600

c: \(38\cdot73+27\cdot38\)

\(=38\left(73+27\right)=38\cdot100=3800\)

Bài 1:

a: \(28\cdot76+23\cdot28-28\cdot13\)

\(=28\left(76+23-13\right)=28\cdot86=2408\)

b: \(39\cdot50+25\cdot39+75\cdot61\)

\(=39\left(50+25\right)+75\cdot61\)

\(=39\cdot75+75\cdot61=75\left(39+61\right)=75\cdot100=7500\)

c: \(32\cdot163+837\cdot32\)

\(=32\left(163+837\right)=32\cdot1000=32000\)

d: \(63+118+37+82\)

=63+37+118+82

=100+200

=300

Câu c:

C = \(9^{2n+1}\) + 1

CM C ⋮ 10

Giải:

9 ≡ -1 (mod 10)

\(9^{2n+1}\) ≡ -1\(^{2n+1}\) (mod 10)

9\(^{2n+1}\) ≡ -1 (mod 10)

1 ≡ 1 (mod 10)

Cộng vế với vế ta có:

9\(^{2n+1}\) + 1 ≡ (-1) + 1 (mod 10)

9\(^{2n+1}\) + 1 ≡ 0 (mod 10)

C = 9\(^{2n+1}\) + 1 ⋮ 10 (đpcm)

\(n^2+n=n\left(n+1\right)\) là tích của hai số tự nhiên liên tiếp

=>\(n^2+n\) chỉ có thể có tận cùng là 0;2;6

=>\(n^2+n+1\) sẽ có tận cùng là 1;3;7

mà \(1995^{2000}\) có chữ số tận cùng là 5

nên \(n^2+n+1\) sẽ không chia hết cho \(1995^{2000}\)

bài 14:

\(a.\left(x-1\right)\cdot100=0\)

\(x-1=0\Rightarrow x=1\)

\(b.200-11x=24\)

\(11x=200-24\)

\(11x=176\)

\(x=\frac{176}{11}=16\)

\(c.165:\left(2x+1\right)=15\) (đkxđ: x khác \(-\frac12)\)

\(2x+1=\frac{165}{15}=11\)

\(2x=11-1=10\)

\(x=\frac{10}{2}=5\)

\(d.375:\left(45-4x\right)=15\) (đkxđ: \(x\ne\frac{45}{4})\)

\(45-4x=\frac{375}{15}=25\)

\(4x=45-25=20\)

\(x=20:4=5\)

bài 15:

giá tiền 125 chiếc điện thoại là:

125 x 2350000=293750000 (đồng)

giá tiền 250 chiếc máy tính bảng là:

250 x 4950000 = 1237500000 (đồng)

tổng số tiền mà cửa hàng phải trả cho số điện thoại và máy tính trên là:

293750000 + 1237500000 = 1531250000 (đồng)

đáp số: 1531250000 đồng

\(1,2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+12\cdot4=3+48=51\)

\(2)5\cdot\left\lbrack\left(85-35:7\right):8+90\right\rbrack-50\)

\(=5\cdot\left\lbrack\left(85-5\right):8+90\right\rbrack-50\)

\(=5\cdot\left(80:8+90\right)-50\)

\(=5\cdot\left(10+90\right)-50=5\cdot100-50\)

\(=500-50=450\)

\(3)2\cdot\left\lbrack\left(7-3^3:3^2\right):2^2+99\right\rbrack-100\)

\(=2\cdot\left\lbrack\left(7-3\right):4+99\right\rbrack-100\)

\(=2\cdot\left(1+99\right)-100=2\cdot100-100\)

\(=200-100=100\)

\(4)2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\)

\(=2^5+5^2\cdot16-3\cdot32=32+25\cdot16-96\)

\(=32+400-96=336\)

\(5)5\cdot2^2\cdot2^3-4\cdot\left(5^8:5^6\right)=5\cdot2^5-4\cdot5^2\)

\(=5\cdot32-4\cdot25=160-100=60\)

\(6)\left(3^5\cdot3^7\right):3^{10}+5\cdot2^4-7^3:7\)

\(=3^{12}:3^{10}+5\cdot16-7^2=9+80-49=40\)

\(7)15:\left(3^5:3^4\right)-2^9:2^7\)

\(=15:3-4=5-4=1\)

\(8)5\cdot3^5:\left(3^8:3^5\right)-2^3\cdot5\)

\(=5\cdot3^2-40=5\cdot9-40=45-40=5\)

\(9)4\left\lbrack\left(3+3^7:3^4\right):10+97\right\rbrack-300\)

\(=4\left\lbrack\left(3+3^3\right):10+97\right\rbrack-300\)

\(=4\left\lbrack\left(3+27\right):10+97\right\rbrack-300\)

\(=4\cdot\left(3+97\right)-300=400-300=100\)

\(10)5\left\lbrack\left(92+2^5:2^2\right):5^2+2^4\right\rbrack-7^2\)

\(=5\left\lbrack\left(92+8\right):25+16\right\rbrack-49\)

\(=5\cdot\left(100:25+16\right)-49\)

\(=5\cdot\left(4+16\right)-49=100-49=51\)

\(11)3^2\cdot\left\lbrack5^2-3\right):11]-2^4+2\cdot10^3\)

\(=9\cdot\left(22:11\right)-16+2000=9\cdot2-16+2000\)

\(=18-16+2000=2002\)

\(12)2^2\cdot5\left\lbrack\left(5^2+2^3\right):11-2\right\rbrack-3^2\cdot2\)

\(=4\cdot5\left\lbrack\left(25+8\right):11-2\right\rbrack-18\)

\(=20\cdot\left(3-2\right)-18=20-18=2\)

\(13)\left(6^{2007}-6^{2006}\right):6^{2006}\)

\(=\left\lbrack6^{2006}\cdot\left(6-1\right)\right\rbrack:6^{2006}=5\)

14) \(\left(5^{2001}-5^{2000}\right):5^{2000}\)

\(=\left\lbrack5^{2000}\cdot\left(5-1\right)\right\rbrack:5^{2000}=4\)

\(15)\left(7^{2005}+7^{2004}\right):7^{2004}\)

\(=\left\lbrack7^{2004}\cdot\left(7+1\right)\right\rbrack:7^{2004}=8\)

\(16)\left(11^{2023}+11_{}^{2022}\right):11^{2022}\)

\(=\left\lbrack11^{2022}\cdot\left(11+1\right)\right\rbrack:11^{2022}=12\)