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4 tháng 10 2018

22 + (x - 3) = 52

4 + (x - 3) = 25

x - 3 = 25 - 4

x - 3 = 21

x = 21 + 3

x = 24

6 tháng 10 2018

22 + ( x - 3 ) = 52

<=> 4 + ( x - 3 ) = 25

<=> x - 3 = 21

<=> x = 24

9x - 2 . 32 = 34

<=>9x - 2 = 34 : 32

<=>9x - 2 = 9

<=>9x = 11

<=> x = 11/9

10x + 22 . 5 = 102

<=> 10x + 4 . 5 = 100

<=> 10x + 4 = 20

<=> 10x = 16

<=> x = 1,6

10 tháng 6 2017

\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\\ =\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\\ =6+6+6+...+6\left(10\text{ số }6\right)\\ =6\cdot10\\ =60\)

\(b,3+\left(-12\right)+13+\left(-22\right)+23+\left(-32\right)+...+93+\left(-102\right)\\ =\left[3+\left(-12\right)\right]+\left[13+\left(-22\right)\right]+\left[23+\left(-32\right)\right]+...+\left[93+\left(-102\right)\right]\\ =\left(-9\right)+\left(-9\right)+\left(-9\right)+...+\left(-9\right)\left(10\text{ số }-9\right)\\ =\left(-9\right)\cdot10\\ =-90\)

10 tháng 6 2017

\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\)

\(=\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\)

\(=6+6+6+...+6\) (10 số 6)

\(=6.10=60\)

\(\)

23 tháng 10 2023

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:

\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)

\(=102-150:\left[18-2\cdot4\right]+1\)

\(=103-\dfrac{150}{18-8}=103-15=88\)

AH
Akai Haruma
Giáo viên
2 tháng 1 2021

Lời giải:

\(B=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2\)

\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2=2^2(1^2+2^2+...+10^2)\)

\(=4A=4.385=1540\)

19 tháng 6 2023

\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)

Gọi 1+2+3+...+10 là P

Số số hạng là: (10 - 1) : 1 +1 = 10 (số)

P = (10+1) . 10 : 2 = 55 

P = 55

Gọi \(1\cdot2+2\cdot3+....+9\cdot10\)  là C

\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)

\(=>A=P+C\\ =>A=55+330\\ A=385\)

b)

\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)

\(\left(1+2^2+3^2+....+10^2\right)=A\)

\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)

22 tháng 8 2023

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(.....\)

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)

\(\Rightarrow B< 1\left(dpcm\right)\)

22 tháng 8 2023

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)

 \(B< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)

 \(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B< 1-\dfrac{1}{10}\)

\(B< \dfrac{9}{10}< 1\)

Vậy \(B< 1\)

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

 \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

......

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )

Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

         \(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)

         \(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)

  .........................

         \(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)

=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

=> D <  1 - \(\dfrac{1}{10}\)

=>D < \(\dfrac{9}{10}\)

=> D < \(\dfrac{10}{10}\)

 Vậy D < 1