25812 + 444565 

= 470377

chúc bn học tốt

25812 + 444565 = 470377

k mk nha bn thân

25812+444565=chet

(-125) + 225 - (-125) + 75 =(-125)+225+125+75=225+75+125-125=300

50

50

chưa chắc

a/19.5

b/0.96

c/1.6

d/-63.952

a) \(A=125\left(-24\right)+24.225\)

\(=24\left[\left(-125\right)+225\right]\)

\(=24.100=2400\)

b) \(B=26\left(-125\right)-125\left(-36\right)\)

\(=125\left[\left(-26\right)-\left(-36\right)\right]\)

\(=125.10=1250\)

c) \(C=25\left(75-45\right)-75\left(45-25\right)\)

\(=25.75-25.45-75.45-75.25\)

\(=\left(25.75-75.25\right)-25.45-75.45\)

\(=-25.45-75.45\)

\(=45\left(-25-75\right)\)

\(=45\left(-100\right)=-4500\)

1) 125-(-75)-326+26

= 125+75-326+26

=200-300

=-100

2) (-5).(-3)³-(-2).5²-75

= -5.-27+2.25-75

= -135+50-75

=-160

Câu1: Tính

1) 125-(-75)-326+26

=125+75-326+26

=125+75-300

=200-300

=-100

2) (-5).(-3)³-(-2).5²-75

=(-5).(-27)-(-50)-75

=(-5).(-27)+50-75

=135+50-75

=110

1)125-(-75)-326+26

=125+75-326+26

=200-300

=-100

2) (-5).(-3)³-(-2).5²-75

=-5.(-27)+2.25-75

=-135+50-75

=-160

A = 125(125 + 150) + 75²

A = 34375 + 75²

A = 34375 + 5652

A = 40000

\(A=125\left(125+150\right)+75^2\)

\(=125^2+125.150+75^2\)

\(=\left(125+75\right)^2\)

\(=200^2=40000\)

2 câu đấy à?

\(125+173+175+27=\left(125+175\right)+\left(173+27\right)=300+200=500\)

\(125\times75+125\times25=125\times\left(75+25\right)=125\times100=12500\)

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 \(n\in\left\{0;1;4;9\right\}\)