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Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
\(a)47-\left[\left(45.2^4-5^2.12\right):14\right]\)
\(=47-\left[\left(45.16-25.12\right):14\right]\)
\(=47-\left[\left(720-300\right):14\right]\)
\(=47-\left[420:14\right]\)
\(=47-30\)
\(=17\)
\(b)50-\left[\left(20-2^3\right):2+34\right]\)
\(=50-\left[\left(20-8\right)\right]:2+34\)
\(=50-\left[12:2+34\right]\)
\(=50-\left[6+34\right]\)
\(=50-40\)
\(=10\)
\(c)10^2-\left[60:\left(5^6:5^4-3,5\right)\right]\)
\(=10^2-\left[60:\left(5^2-3,5\right)\right]\)
\(=10^2-\left[60:\left(25-3,5\right)\right]\)
\(=10^2-\left[60:21,5\right]\)
\(=100-\dfrac{120}{43}\)
\(=\dfrac{4180}{43}\)
\(d)50-\left[\left(50-2^3.5\right):2+3\right]\)
\(=50-\left[\left(50-8.5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left[10:2+3\right]\)
\(=50-5+3\)
\(=50-8\)
\(=42\)
\(e)10-\left[\left(8^2-48\right).5+\left(2^3.10+8\right)\right]:28\)
\(=10-\left[\left(64-48\right).5+\left(8.10+8\right)\right]:28\)
\(=10-\left[16.5+\left(80+8\right)\right]:28\)
\(=10-\left[80+88\right]:28\)
\(=10-168:28\)
\(=10-6\)
\(=4\)
\(f)8697-\left[3^7:3^5+2.\left(13-3\right)\right]\)
\(=8697-\left[3^2+2.\left(13-3\right)\right]\)
\(=8697-\left[9+2.10\right]\)
\(=8697-9+20\)
\(=8697-29\)
\(=8668\)
\(g)2011+5\left[300-\left(17-7\right)^2\right]\)
\(=2011+5.\left[300-10^2\right]\)
\(=2011+5.\left[300-100\right]\)
\(=2011+5.200\)
\(=2011+1000\)
\(=3011\)
\(h)695-\left[200+\left(11-1\right)^2\right]\)
\(=695-\left[200+10^2\right]\)
\(=695-200+100\)
\(=695-300\)
\(=395\)
\(i)129-5\left[29-\left(6-1\right)^2\right]\)
\(=129-5\left[29-5^2\right]\)
\(=129-5\left[29-25\right]\)
\(=129-5.4\)
\(=129-20\)
\(=109\)
c)
\(4\left(3x-4\right)-2=18\)
<=> \(12x-16-2=18\)
<=> \(12x=36\)
<=> \(x=3\)
Vậy x=3
d)
\(\left(3x-10\right):10=50\)
<=> \(3x-10=500\)
<=> \(3x=510\)
<=> x= \(170\)
Vậy x= 170
f)
\(x-\left[42+\left(-25\right)\right]=-8\)
<=> \(x-17=-8\)
<=> x= \(9\)
Vậy x=9
h)
\(x+5=20-\left(12-7\right)\)
<=> \(x+5=15\)
<=> \(x=10\)
Vậy x= 10
k)
\(\left|x-5\right|=7-\left(-3\right)\)
<=> \(\left|x-5\right|=10\)
* Với \(x>=5\) ; ta được:
\(x-5=10\)
<=> x= 15 (thoả mãn điều kiện )
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=10\)
<=> \(-x+5=10\)
<=> \(-x=5\)
<=> \(x=-5\) (thoả mãn điều kiện)
Vậy x=15 ; x= -5
i)
\(\left|x-5\right|=\left|7\right|\)
<=> \(\left|x-5\right|=7\)
*Với \(x>=5\) ; ta được:
\(x-5=7\)
<=> \(x=12\) (thoả mãn)
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=7\)
<=> \(-x=2\)
<=> \(x=-2\) (thoả mãn)
Vậy x= 12; x= -2
m)
\(2^{x+1}.2^{2009}=2^{2010}\)
<=> \(2^{x+1+2009}=2^{2010}\)
<=> \(2^{x+2010}=2^{2010}\)
=> \(x+2010=2010\)
=> \(x=0\)
Vậy x=0
n)
\(10-2x=25-3x\)
<=>\(x=15\)
Vậy x=15
a) 12 : { 400 : [500 – (125 + 25 . 7)]}
=12 : { 400 : [500 – (125 + 175)]}
=12 : [ 400 : [500 – 300)]
=12 : (400:200)
=12:2=6
b) 5 . 22 – 18 : 3
=5.4-6
=20-6=14
c) 18 : 3 + 182 + 3.(51 : 17)
=5+182+3.3
=187+9=196
d) 25 . 8 – 12.5 + 170 : 17 – 8
=200-60+10-8
=142
e) 2.52+ 3: 710 – 54: 33
=2.25+3:1-54:27
=50+3-2
=51
f) 189 + 73 + 211 + 127
=(189+211)+(73+127)
=400+200=600
g) 375 : {32 – [ 4 + (5. 32– 42)]} – 14 )
=375 : {32 – [ 4 + (160– 42)]} – 14 )
=375 : [32 – ( 4 + 118) – 14 ]
=375:(32-122-14)
=375:-104
=-375/104
h) (52022 + 52021) : 52021
=(52022:52021)+(52021:52021)
=5+1=6
a) Ta có: \(14-\left(5-8\right)^3+\left(-3\right)\cdot5\)
\(=14+27-15=26\)
b) Ta có: \(-7\cdot15+7\cdot\left(-35\right)+\left(-1\right)^{2019}\)
\(=-7\left(15+35\right)-1=-350-1=-351\)
c) Ta có: \(-18\cdot32+\left(-18\right)\cdot45+77\cdot\left(-32\right)-77\cdot50\)
\(=-18\left(32+45\right)+77\left(-32-50\right)\)
\(=-18\cdot77+77\cdot\left(-82\right)=77\left(-18-82\right)=77\cdot\left(-100\right)=-7700\)
d) Ta có: \(104-4\cdot\left[-5\cdot8+\left(7-10\right)^3\right]\)
\(=104-4\left[-40-27\right]=104-4\cdot\left(-67\right)=104+268=372\)
Tính:
\(A=\left(-2\right)^3.\left(-5\right)^2-41.\left(-8\right)\)
\(=\left(-8\right).25-41.\left(-8\right)\)
\(=\left(-8\right).\left(25-41\right)\)
\(=\left(-8\right).\left(-16\right)\)
\(=128\)
\(B=2018-2018:\left(4^2-3^2\right)\)
\(=2018-2018:\left(16-9\right)\)
\(=2018-2018:5\)
\(=2018-403,6\)
\(=1614,4\)
\(C=\left(-5\right).\left(-25\right).\left(-50\right)+5.25.50\)
\(=\left(-6250\right)+6250\)
\(=0\)
a. 4 . 5 + 28 : 7 - 620 : 618
= 20 + 4 - 62
= 20 + 4 - 36
= .... ( tự làm )
b. 718 : 716 + 22 . 33
= 72 + 25
= 49 + 32
= ...
c. 59 : 57 + 12 . 3 + 70
= 52 + 36 + 0
= ...
d. 205 - [ 1200 - ( 42 - 2 . 3 )3 ]
= 205 - [ 1200 - 103 ]
= 205 - [ 1200 - 100 ]
= ...
e. 695 - [ 200 - ( 11 - 1 )2 ]
= 695 - [ 200 - 102 ]
= 695 - [ 200 - 100 ]
= ...
#A?m?
sai hết r bạn ơi
phải là
a)28
b)157
c)62
d)5
e)598
ko có số âm đâu
a: \(20-\left[30-\left(5-1\right)^2\right]\)
\(=20-\left[30-4^2\right]\)
\(=20-14=6\)
b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)
\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)
\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)
c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)
\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)
\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)
d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)
\(=411-\left[\dfrac{110}{5}-4\right]\)
=410-22+4
=410-18
=392
e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)
\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)
\(=450-5\cdot\left[9\cdot8-12\right]+18\)
=468-5*60
=468-300
=168
f:
\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)
\(=102-150:\left[18-2\cdot4\right]+1\)
\(=103-\dfrac{150}{18-8}=103-15=88\)