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26 tháng 7 2017

a) 5 x 4 + 117 = 20 + 117

                   = 137

b) 200 : 2 – 75 = 100 – 75

                   = 25

24 tháng 12 2023

a: \(12:\dfrac{3}{4}+75\%\cdot\dfrac{1}{2}-16\cdot50\%\)

\(=12\cdot\dfrac{4}{3}+\dfrac{3}{4}\cdot\dfrac{1}{2}-16\cdot0,5\)

\(=16-8+\dfrac{3}{8}\)

\(=8,375\)

b: \(75\%\cdot4+22,5:3-1\dfrac{3}{5}\)

\(=0,75\cdot4+7,5-1,6\)

\(=3+7,5-1,6\)

=10,5-1,6

=8,9

20 tháng 5 2018

ai nhanh nhất mình k cho

20 tháng 5 2018

Bài 1:

\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)

\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)

\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(A=\frac{9}{11}\)

Bài 2 :

2)  Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15

= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15

Mẫu số =  11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19  + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17  

=  (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x  17 = (1+27+125+729) x 13 x 15 x 17 

\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)

\(=\frac{11}{17}\)

\(=\frac{1111}{1717}=B\)

Vậy \(A=B\)

13 tháng 6 2015

1)\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2.\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{11}\)


 

13 tháng 6 2015

2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15

= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15

Mẫu số =  11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19  + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 

lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 

=  (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x  17 = (1+27+125+729) x 13 x 15 x 17

=> \(A

28 tháng 10 2023

1:

a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)

b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)

=>2x-1>0

=>x>1/2

2:

a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)

\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)

\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)

\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)

\(=45\sqrt{2}-19\sqrt{5}\)

b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)

\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)

\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)

Giải:

a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\) 

=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\) 

=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\) 

=\(\dfrac{7}{5}+\dfrac{-1}{4}\) 

=\(\dfrac{23}{20}\) 

b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\) 

=\(\left[0+0\right]:\dfrac{2017}{2018}\) 

=0\(:\dfrac{2017}{2018}\) 

=0

c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)

=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) 

=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10

11 tháng 11 2021

a)

=20x - 5y

=4x-y

11 tháng 11 2021

b: \(=x^3+4x^2+2x^2+8x-3x-12\)

\(=x^3+6x^2+5x-12\)

29 tháng 4 2022

a, \(1+\dfrac{3}{4}=\dfrac{7}{4}\)

b, \(\dfrac{4}{5}-\dfrac{3}{8}=\dfrac{32-15}{40}=\dfrac{17}{40}\)

c, \(1:\dfrac{2}{3}=\dfrac{1.3}{2}=\dfrac{3}{2}\)

d, \(\dfrac{2}{5}.\dfrac{5}{2}=1\)

29 tháng 4 2022

a)1 + 3/4=7/4

   b)4/5 -3/8=17/40

  c)1 : 2/3=3/2

  d)2/5 x 5/2=1

21 tháng 12 2023

Bài 1:

a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)

\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)

\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)

\(=10\sqrt{2}-16\sqrt{3}\)

b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)

\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)

\(=3-\sqrt{6}+\sqrt{6}-1\)

=3-1=2

c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)

\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)

\(=\sqrt{15}+4-\sqrt{15}=4\)

d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)

\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)

\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)

Bài 2:

Vẽ đồ thị:

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Phương trình hoành độ giao điểm là:

\(\dfrac{1}{2}x-4=-3x+3\)

=>\(\dfrac{1}{2}x+3x=3+4\)

=>\(\dfrac{7}{2}x=7\)

=>x=2

Thay x=2 vào y=-3x+3, ta được:

\(y=-3\cdot2+3=-3\)

Vậy: (d1) cắt (d2) tại A(2;-3)