có bị sai đề ko bạn sao vài chỗ ko có dấu cộng trừ nhân chia vậy??

12. (-47) + 12. (-19) + (-66). 88

= 12. [(-47) + (-19)] + (-66). 88

= 12. (-66) + (-66). 88

= (-66). (12 + 88)

= (-66). 100

= -6600

A = 12.(-47) + 12.(-19) + (-66).88

A = -12.(47 + 19)  - 66.88

A = -12.66 - 66.88

A = -66.(12 + 88)

A = -66. 100

A = - 6600

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 \(n\in\left\{0;1;4;9\right\}\)

12 + 88 + 34 + 66

= ( 12 + 88 )+(34 + 66 )

= 100 +100

= 200

=(12+88)+(34+66)

=100 +100

=200

đúng nha bn kik mình nhé

Tính nhanh:

12 + 88 + 34 + 66 

= ( 12 + 88 ) + ( 34 + 66 )

= 100 + 100

= 200

Vậy giá trị biểu thức trên là : 200

e. = 12 - 25 + 38 + 25 - 38 + 88  - 250 

    = ( -25 + 25) + (38 - 38) + ( 12 + 88) - 250

   = -150

f. = (-12).44 + 44.(-88)

 = 44 ( -12 -88 )

= 44 . 100

= 4400

g. = -48 + 54 -77 -24 + 48 + 77

= (-48 + 48) + (-77+77) + (54 -24)

= 30

h. = 11.(-42) + 11.(-26) + 42.11 + 42.26 -26.11 - 42.26

 = \([11.\left(-42\right)+11.42]+\left[11.\left(-26\right)-26.11\right]+[42.26-42.26]\)

= 11. (-52)

= - 572

\(e.-\left(-12+25-38\right)+\left(25-38+88\right)-250\\ =12-25+38+25-38+88-250\\ =\left(25-25\right)+\left(38-38\right)+\left(12+88\right)-250\\ =0+0+100-250=-150\)

\(f.\left(-12\right)\cdot\left(67-23\right)+44\cdot\left(-88\right)\\ =\left(-12\right)\cdot44+44\cdot\left(-88\right)\\ =44\cdot\left[-12+\left(-88\right)\right]\\ =44\cdot\left(-100\right)=-4400\)

\(g.\left(-48+54-77\right)-\left(24-48-77\right)\\ =-48+54-77-24+48+77\\ =\left(48-48\right)+\left(54-24\right)+\left(77-77\right)\\ =0+30+0=30\)

\(h.11\cdot\left(-42-26\right)+42\cdot\left(11+26\right)-26\cdot11-42\cdot26\\ =11\cdot\left(-68\right)+42\cdot37-26\cdot11-42\cdot26\\ =-748+1554-286-1092=-572\)

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

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

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

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

d: \(-37\cdot\left(-19\right)+37\left(-19\right)\)

\(=-19\left(-37+37\right)\)

\(=-19\cdot0=0\)

e: \(\left(29-9\right)\cdot\left(-9\right)+\left(-13-7\right)\cdot21\)

\(=20\cdot\left(-9\right)+\left(-20\right)\cdot21\)

\(=-20\left(9+21\right)\)

\(=-20\cdot30=-600\)

h: \(12\cdot\left(-47\right)+12\left(-19\right)+\left(-66\right)\cdot88\)

\(=12\left(-47-19\right)+\left(-66\right)\cdot88\)

\(=\left(-66\right)\cdot12+\left(-66\right)\cdot88\)

\(=-66\left(12+88\right)\)

\(=-66\cdot100=-6600\)

\(12\cdot\left(-47\right)+12\cdot\left(-19\right)+\left(-66\right)\cdot88\)

\(=12\left(-47-19\right)+\left(-66\right)\cdot88\)

\(=-66\cdot12+88\cdot\left(-66\right)\)

\(=-66\left(12+88\right)\)

\(=-66\cdot100=-6600\)