Giải:

( 12 + 77 + 34 + 23 + 88 + 45 + 55 + 66 ) . ( 45200 - 1808 .25 )

=( 12 + 77 + 34 + 23 + 88 + 45 + 55 + 66 ) . ( 45200 - 45200 )

=( 12 + 77 + 34 + 23 + 88 + 45 + 55 + 66 ) . 0

=0

Học tốt!!!

(12+77+34+23+88+45+55+66)* (45200-1808*25)

=(12+77+34+23+88+45+55+66)* (45200-45200)

=(12+17+34+23+88+45+55+66)* 0

=0

\(\left(12+77+34+23+88+45+55+66\right)\times\left(45200-1808\times25\right)\)

\(=\left(12+77+34+23+88+45+55+66\right)\times\left(45200-45200\right)\)

\(=\left(12+77+34+23+88+45+55+66\right)\times0\)

\(=0.\)

\(=\left[66+34+77+23+12+88+45+55\right]\times\left[1808\times\left(25-25\right)\right]\\ =0\)

(45+75)*66+(75+45)*34

=(45+75)*(66+34)

=120*100

=12000

11+22+33+44+55+66+77+88+99:11

=11+22+33+44+55+66+77+88+9

=(11+9)+(22+88)+(33+77)+(44+66)+55

=20+110*3+55

=20+330+55

=405

Ta có :

a, (45 + 75) x 66 + (75 + 45) x 34

=  (45 + 75) x (66 + 34)

=   120  x 100

=  12000

11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 : 11

= 33 + 77 + 121 + 165 + 9

= 405

\(11+22+33+44+55+66+77+88+99:11\)

\(\left(11+99\right)+\left(22+88\right)+\left(33+77\right)+\left(44+66\right)+55:11\)

\(110+110+110+110+55:11\)

\(440+55:11\)

\(445\)

Bài 2:

a: \(17-x=3\)

=>\(x=17-3\)

=>x=14(nhận)

b: \(2\cdot\left(x-1\right):3=6\)

=>\(2\left(x-1\right)=6\cdot3=18\)

=>x-1=18/2=9

=>x=9+1=10(nhận)

c: \(x+\left(-2\right)=\left(-11\right)+7\)

=>\(x-2=-4\)

=>\(x=-4+2=-2\left(loại\right)\)

d: \(\left(x-1\right)^2-5=20\)

=>\(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Câu 3:

a: Đặt *=a

\(\overline{57a3}⋮9\)

=>\(5+7+a+3⋮9\)

=>\(a+15⋮3\)

mà 0<=a<=9

nên a=3

=>*=a

b: \(A=123\cdot7+8+9\)

123*7 là số lẻ

9 là số lẻ

=>123*7+9 chia hết cho 2

mà 8 chia hết cho 2

nên \(A=123\cdot7+9+8⋮2\)

\(123\cdot7⋮3;9⋮3;8⋮̸3\)

=>\(A=123\cdot7+9+8⋮̸3\)

c: \(B=3\cdot5\cdot7+10^{50}\)

\(=5\cdot3\cdot7+5\cdot5^{49}\cdot2^{49}\)

\(=5\left(3\cdot7+5^{49}\cdot2^{49}\right)⋮5\)

=>B là hợp số

con chóa này

*Note: Tích của một số với 0 luôn bằng 0.

a) \(\left(1763+7634+23\cdot764\right)\cdot\left(346\cdot45-173\cdot90\right)\)

\(=\left(1763+7634+23\cdot764\right)\cdot\left(346\cdot45-173\cdot45\cdot2\right)\)

\(=\left(1763+7634+23\cdot764\right)\cdot\left(346\cdot45-346\cdot45\right)\)

\(=\left(1763+7634+23\cdot764\right)\cdot0\)

\(=0\)

b) \(N=10\cdot12\cdot14\cdot16\cdot\left(88\cdot17-44\cdot34\right)\)

\(N=10\cdot12\cdot14\cdot16\cdot\left(88\cdot17-44\cdot17\cdot2\right)\)

\(N=10\cdot12\cdot14\cdot16\cdot\left(88\cdot17-88\cdot17\right)\)

\(N=10\cdot12\cdot14\cdot16\cdot0\)

\(N=0\)

c) \(\left(80\cdot96+40\cdot1808\right):1000\)

\(=\left(80\cdot96+40\cdot904\cdot2\right):1000\)

\(=\left(80\cdot96+80\cdot904\right):1000\)

\(=\left[80\cdot\left(96+904\right)\right]:1000\)

\(=80\cdot1000:1000\)

\(=80\)

\(a)\left(80\times96+40\times1808\right):1000\)

\(=\left(80\times96+40\times2\times904\right):1000\)

\(=\left(80\times96+80\times904\right):1000\)

\(=80\times\left(96+904\right):1000\)

\(=80\times1000:1000\)

\(=80\)

\(b)\) em xem lại đề xem có sai gì không em.

\(c)10\times12\times14\times16\times\left(88\times17-44\times34\right)\)

\(=10\times12\times14\times16\times\left(44\times2\times17-44\times2\times17\right)\)

\(=10\times12\times14\times16\times0\)

\(=0\)

34 < 50     47 > 45     55 < 66

78 > 69     81< 82     44 > 33

72 < 81     95 > 90     77 < 99

62 = 62     61 < 63     88 > 22

\(\dfrac{x-9}{55}+\dfrac{x-10}{66}=\dfrac{x-11}{77}+\dfrac{x-12}{88}\\ \Leftrightarrow\left(\dfrac{x-9}{55}+\dfrac{1}{11}\right)+\left(\dfrac{x-10}{66}+\dfrac{1}{11}\right)-\left(\dfrac{x-11}{77}+\dfrac{1}{11}\right)-\left(\dfrac{x-12}{88}+\dfrac{1}{11}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-9}{55}+\dfrac{5}{55}\right)+\left(\dfrac{x-10}{66}+\dfrac{6}{66}\right)-\left(\dfrac{x-11}{77}+\dfrac{7}{77}\right)-\left(\dfrac{x-12}{88}+\dfrac{8}{88}\right)=0\)

\(\Leftrightarrow\dfrac{x-4}{55}+\dfrac{x-4}{66}-\dfrac{x-4}{77}-\dfrac{x-4}{88}=0\)

\(\Leftrightarrow\left(x-4\right)\left(\dfrac{1}{55}+\dfrac{1}{66}-\dfrac{1}{77}-\dfrac{1}{88}\right)=0\\ \Leftrightarrow x=4\left(vì.\dfrac{1}{55}+\dfrac{1}{66}-\dfrac{1}{77}-\dfrac{1}{88}\ne0\right)\)

1.

a) \(125+80+375+220\)

\(=\left(125+375\right)+\left(80+220\right)\)

\(=500+300\)

\(=800\)

b) \(25.11.8.4.125\)

\(=\left(25.4\right).\left(8.125\right).11\)

\(=100.1000.11\)

\(=1100000\)

c) \(18.74+18.27-18\)

\(=18.\left(74+27-1\right)\)

\(=18.100\)

\(=1800\)

d) \(75.23+75.77-500\)

\(=75.\left(23+77\right)-500\)

\(=75.100-500\)

\(=7500-500\)

\(=7000\)

e) \(150:\left[25.\left(18-4^2\right)\right]\)

\(=150:\left[25.\left(18-16\right)\right]\)

\(=150:\left[25.2\right]\)

\(=150:50\)

\(=3\)

f) \(125.23.2.8.50\)

\(=\left(125.8\right).\left(2.50\right).23\)

\(=1000.100.23\)

\(=2300000\)

g) \(235+88+165+12\)

\(=\left(235+165\right)+\left(88+12\right)\)

\(=400+100\)

\(=500\)

h) \(71.32+71.68-1100\)

\(=71.\left(32+68\right)-1100\)

\(=71.100-1100\)

\(=7100-1100\)

\(=6000\)

i) \(6^7:6^6+4^3.2-24^0\)

\(=6+64.2-1\)

\(=6+128-1\)

\(=133\)

j) \(546-6.\left[158:\left(30+7^2\right)\right]\)

\(=546-6.\left[158:\left(30+49\right)\right]\)

\(=546-6.\left[158:79\right]\)

\(=546-6.2\)

\(=546-12\)

\(=534\)

k) \(268+147+132+253\)

\(=\left(268+132\right)+\left(147+253\right)\)

\(=400+400\)

\(=800\)

2. 

a) \(7x=707\)

\(x=707:7\)

\(x=101\)

b) \(6x+5=47\)

\(6x=47-5\)

\(6x=42\)

\(x=42:6\)

\(x=7\)

c) \(35:\left(x+1\right)=7\)

\(x+1=35:7\)

\(x+1=5\)

\(x=5-1\)

\(x=4\)

d) \(12+\left(29-3x\right)=35\)

\(29-3x=35-12\)

\(29-3x=23\)

\(3x=29-23\)

\(3x=6\)

\(x=6:3\)

\(x=2\)

e) \(2.4^x+10^1=138\)

\(2.4^x=138-10\)

\(2.4^x=128\)

\(4^x=128:2\)

\(4^x=64=4^3\)

\(\Rightarrow x=3\)

f) \(7x=140\)

\(x=140:7\)

\(x=20\)

g) \(7\left(15-x\right)+14=84\)

\(7\left(15-x\right)=84-14\)

\(7\left(15-x\right)=70\)

\(15-x=70:7\)

\(15-x=10\)

\(x=15-10\)

\(x=5\)

h) \(3^x.5-15=390\)

\(3^x.5=390+15\)

\(3^x.5=405\)

\(3^x=405:5\)

\(3^x=81=3^4\)

\(\Rightarrow x=4\)

3.

Ta có:

\(\Rightarrow n+11⋮n+1\)

\(\Rightarrow n+1+10⋮n+1\)

\(\Rightarrow10⋮n+1\) hay \(n+1\inƯ\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Rightarrow\) \(n\in\left\{-11;-6;-3;-2;0;1;4;9\right\}\)

Vì \(n\in N\) nên \(n\in\left\{0;1;4;9\right\}\)