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)

  23.125 + 78.125 -125

=125.(23+78-1)

=125.100

=12500

b)

  37.46+18.54-17.46+54.2

=46.(37-17)+54.(18+2)

=46.20+54.20

=(46+54).20

=100.20

=2000

c)

  3.2²-2³:2+2²

=3.2²-2²+2²

=3.2²-(2²-2²)

=3.2²

=3.4

=12

d)

  7³.9+3².7⁴-45.539

=7³.9+9.7³.7-9.5.7².11

=9.7³.(1+7)-9.7².(5.11)

=9.7².7.8-9.7².55

=9.7².(7.8-55)

=9.49.(56-55)

=441.1

=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, 34-29+(-14)+129-(-114)
= 34-29-14+129+114
= (34-14)+(-29+129)+114
= 20 + 100 + 114
= 234

câu C đề sai hay sao í

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

Đáp án là A