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80-[4.52 .3.22 ]
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1: \(23+\left(-13\right)+\left(-50\right)\)
\(=23-13-50\)
=10-50
=-40
2: \(-5+15+\left(-123\right)\)
\(=\left(-5+15\right)-123\)
=10-123
=-113
3: \(5871:\left\{928-\left[\left(-82+247\right)\right]\right\}\cdot5\)
\(=5871-\left\{928+82-247\right\}\cdot5\)
\(=5871-763\cdot5=5871-3815=2056\)
4: \(40-\left(4\cdot5^2-3\cdot2^3\right)\)
\(=40-4\cdot5^2+3\cdot2^3\)
\(=40-4\cdot25+3\cdot8\)
=40-100+24
=64-100
=-36
5: \(6^2\cdot5-7^2+149\)
\(=36\cdot5-49+149\)
\(=180+149-49\)
=180+100
=280
6: \(-210:\left[16+3\cdot\left(6+3\cdot2^2\right)\right]+\left(-2022\right)\)
\(=-210:\left[16+3\cdot\left(6+3\cdot4\right)\right]+\left(-2022\right)\)
\(=-210:\left[16+3\cdot18\right]+\left(-2022\right)\)
\(=-210:70-2022\)
=-3-2022
=-2025
7: \(5\cdot2^3+7^{11}:7^9-2023^0\cdot1^8\)
\(=5\cdot8+7^2-1\)
=40+49-1
=39+49
=88
8: \(400:\left\{5\cdot\left[360-\left(290+2\cdot5^2\right)\right]\right\}\)
\(=400:\left\{5\cdot\left[360-290-2\cdot25\right]\right\}\)
\(=400:\left\{5\cdot20\right\}\)
\(=\dfrac{400}{100}=4\)
9: \(75-\left(3\cdot5^2\right)-4\cdot5^3\)
\(=75-3\cdot25-4\cdot5^3\)
=-4*125
=-500
\(a,23\cdot125+78\cdot125-125\\ =125\cdot\left(23+78-1\right)\\ =125\cdot100\\ =12500\\ b,37\cdot46+18\cdot54-17\cdot46+54\cdot2\\ =46\cdot\left(37-17\right)+54\cdot\left(18+2\right)\\ =46\cdot20+54\cdot20\\ =20\cdot\left(46+54\right)\\ =20\cdot100\\ =2000\\ b,3\cdot2^2-2^3:2+2^2\\ =3\cdot4-4+4\\ =12\\ d,7^3\cdot9+3^2\cdot7^4\\ =7^3\cdot9+7^4\cdot9\\ =7^3\cdot9\cdot\left(1+7\right)\\ =343\cdot9\cdot8\\ =24696\)
a) \(23\cdot125+78\cdot125-125\)
\(=125\cdot\left(23+78-1\right)\)
\(=125\cdot100\)
\(=12500\)
b) \(37\cdot46+18\cdot54-17\cdot46+54\cdot2\)
\(=54\cdot\left(18+2\right)+46\cdot\left(37-17\right)\)
\(=54\cdot20+46\cdot20\)
\(=54\cdot\left(20+46\right)\)
\(=20\cdot100\)
\(=2000\)
c) \(3\cdot2^2-2^3:2+2^2\)
\(=3\cdot2^2-2^2+2^2\)
\(=3\cdot2^2\)
\(=3\cdot4\)
\(=12\)
d) \(7^3\cdot9+3^2\cdot7^4\)
\(=7^3\cdot3^2+3^2\cdot7^4\)
\(=7^3\cdot3^2\cdot\left(1+7\right)\)
\(=24696\)
3:
a: \(\dfrac{\left(x-3\right)}{5}=6^2-2^3\cdot4\)
=>\(\dfrac{x-3}{5}=36-8\cdot4=4\)
=>x-3=20
=>x=23
b: \(3^{x+2}+5\cdot2^3=47+\dfrac{18}{4^2-7}\)
=>\(3^{x+2}+5\cdot8=47+\dfrac{18}{16-7}=49\)
=>\(3^{x+2}=9\)
=>x+2=2
=>x=0
c: \(2^{x+1}-2^x=8^2\)
=>\(2^x\cdot2-2^x=2^6\)
=>\(2^x=2^6\)
=>x=6
d: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\cdot x^2=99\)
=>\(x^2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=99\)
=>\(x^2\cdot\dfrac{99}{100}=99\)
=>\(x^2=100\)
=>\(\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
e: \(\left(2x-3\right)^7=\left(2x-3\right)^5\)
=>\(\left(2x-3\right)^5\left[\left(2x-3\right)^2-1\right]=0\)
=>\(\left(2x-3\right)^5\cdot\left(2x-3-1\right)\left(2x-3+1\right)=0\)
=>\(\left(2x-3\right)^5\left(2x-4\right)\left(2x-2\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
f: \(\left(x-2\right)^{10}=\left(x-2\right)^8\)
=>\(\left(x-2\right)^8\left[\left(x-2\right)^2-1\right]=0\)
=>\(\left(x-2\right)^8\left(x-2-1\right)\left(x-2+1\right)=0\)
=>\(\left(x-2\right)^8\cdot\left(x-3\right)\left(x-1\right)=0\)
=>\(x\in\left\{2;3;1\right\}\)
a) 23.125+78.125-125
= (23+78-1).125
= 100.125
= 12500
b) 37.46+18.54-17.46+54.2
= (37-17).46+(18+2).54
= 20.46+20.54
= 20.(46+54)
= 20.100
= 2000
c) 3.22-23:2+22
= 3.22-22+22
= 3.22
= 3.4
= 12
d) 73.9+32.74-45.539
= 73.9+9.74-5.9.72.11
= 9.(73+74)-5.9.72.11
=9.2744-9.2695
= 9.(2744-2695)
= 9.49
= 441
a/
=125(23+78-1)=125.100=12500
b/
=46(37-17)+54(18+2)
=20(46+54)=20.100=2000
c/
=3.22-22+22=3.4=12
d/
=73.32+32.74-5.32.72.11=
=32.72(7+72-55)=32.72=441
a) $34-29+(-14)+129-(-114)$
$=34-29-14+129+114$
$=34+(129-29)+(114-14)$
$=34+100+100$
$=234$
b) $27\cdot(-35)+(-27)\cdot64-27$
$=(-27)\cdot35+(-27)\cdot64+(-27)\cdot1$
$=(-27)\cdot(35+64+1)$
$=(-27)\cdot(99+1)$
$=(-27)\cdot100$
$=-2700$
c) $160-(4\cdot5^2-3\cdot2^3)\cdot2-(-12)$
$=160-(4\cdot25-3\cdot8)\cdot2+12$
$=160-(100-24)\cdot2+12$
$=160-76\cdot2+12$
$=160-152+12$
$=160-(152-12)$
$=160-140$
$=20$
a, 34 - 29 + (-14) + 129 - (-114)
= 34 + (129 - 29) + (114 - 14)
= 34 + 100 + 100
= 234
b, 27.(-35) + (-27).64 - 27
= (-27).(35 + 64 + 1)
= - 27 . 100
= - 2700
c, 160 - (4.52 - 3.23).2 - (-12)
= 160 - (4.25 - 3.8).2 + 12
= 160 - (100 - 24).2 + 12
= 160 - 76.2 + 12
= 160 - 152 + 12
= 8 + 12
= 20
a)=(4.25).37
=100.37
=3700
b)=(132+868)+(763+237)+29
=1000+1000+29
=2029
c)=72+[131-1.20]
=49+[131-20]
=49+111
=160
Đáp án A
y = x + x 2 + x 3 + ... + x 2018 = x 1 − x 2018 1 − x y ' = 1 + 2 x + 3 x 2 + ... + 2018 x 2017 = 2018 x 2019 − 2019 x 2018 + 1 ( 1 − x ) 2 y ' ( 2 ) = 1 + 2.2 + 3.2 2 + ... + 2018.2 2017 = 2017.2 2018 + 1
\(80-\left(4.5^2.3.2^2\right)\)
\(=80-\left(4.5.5.3.2.2\right)\)
\(=80-\left(\left[5.2\right].\left[5.2\right].\left[4.3\right]\right)\)
\(=80-\left(10.10.12\right)\)
\(=80-\left(10.120\right)\)
\(=80-1200\)
\(=-1120\)
80-[4.5^2 .3.2^2 ]
=80-[4.25.3.4]
=80-[100.12]
=80-1200=-1120