Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1: Đặt A=1 x 2 x 3 + 2 x 3 x 4 + ... + 111 x 112 x 113
⇒4A= 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5-1)+ ... + 111 x 112 x 113 x (114-110)
⇒4A= 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4+ ... + 111 x 112 x 113 x 114 -110 x 111 x 112 x 113
⇒4A=111 x 112 x 113 x 114
⇒A=\(\frac{\text{111 x 112 x 113 x 114}}{4}\)
Câu 2:
Tương tự câu 1, bạn nhân A với 3 là sẽ tính được
Bạn xem lại đề câu a) cho rõ lại
Câu b) Tại x=2013 thì B=x2013-(x+1)x2012+(x+1)x2011-(x+1)x2010+...-(x+1)x2+(x+1)x-1
= x2013-x2013-x2012+x2012+x2011-x2011-x2010+..-x3 - x2+x2+x-1
= x-1 = 2012
\(h\left(x\right)+f\left(x\right)-g\left(x\right)=-2x^2-x+9\)
\(h\left(x\right)+\left(-5x^4+x^2-2x+6\right)-\left(-5x^4+x^3+3x^2-3\right)=-2x^2-x+9\)
\(h\left(x\right)-5x^4+x^2-2x+6+5x^4-x^3-3x^2-3=-2x^2-x+9\)
\(h\left(x\right)-\left(5x^4-5x^4\right)+\left(x^2-3x^2\right)-x^3-2x+\left(6-3\right)=-2x^2-x+9\)
\(h\left(x\right)-0-2x^2-x^3-2x+3=-2x^2-x+9\)
\(h\left(x\right)-x^3-2x^2-2x+3=-2x^2-x+9\)
\(h\left(x\right)+\left(-x^3-2x^2-2x+3\right)=-2x^2-x+9\)
\(h\left(x\right)=\left(-2x^2-x+9\right)-\left(-x^3-2x^2-2x+3\right)\)
\(h\left(x\right)=-2x^2-x+9+x^3+2x^2+2x-3\)
\(h\left(x\right)=\left(-2x^2+2x^2\right)-\left(x-2x\right)+\left(9-3\right)+x^3\)
\(h\left(x\right)=0+x+6+x^3\)
\(h\left(x\right)=x^3+x+6\)
d) Ta có : h(x) + f(x) - g(x) = -2x2 - x + 9
<=> h(x) = -2x2 - x + 9 - f(x) + g(x)
<=> h(x) = -2x2 - x + 9 - x2 + 2x + 5x4 - 6 + x3 - 5x4 + 3x2 - 3
<=> h(x) = x3 + x.
Vậy h(x) = x3 + x
A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
\(f\left(x\right)-g\left(x\right)=5x^2-2x+5-\left(5x^2-6x-\frac{1}{3}\right)\)
= \(5x^2-2x+5-5x^2+6x+\frac{1}{3}\)
=\(4x+\frac{16}{3}\)
Câu 1 : Đặt A = 1.2.3 + 2.3.4 + ... + 111.112.113
=> 4A = 1.2.3.4 + 2.3.4.4 + ... + 111.112.113.4
= 1.2.3.4 + 2.3.4.(5 - 1) + .... + 111.112.113.(114 - 110)
= 1.2.34 + 2.3.4.5 - 1.2.3.4 + ... + 111.112.113.114 - 110.111.112.113
= 111.112.113.114
=> A = 111.113.114.28 = 40 037 256
Câu 2 Đặt A = 1.2 + 2.3 + 3.4 + ... + 277.278
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 277.278.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 277.278.(279 - 276)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 277.278.279 - 276.277.278
= 277.278.279
=> A = 7161558
3) Đặt A = 1.4 + 2.5 + ... + 277.280
= 1.(2 + 2) + 2.(2 + 3) + ... + 277.(278 + 2)
= (1.2 + 2.3 + .... + 277.278) + 2(1 + 2 + .... 277)
Đặt B = 1.2 + 2.3 + .... + 277.278
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 277.278.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 277.278.(279 - 276)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 277.278.279 - 276.277.278
= 277.278.279
=> B = 7161558
Khi đó A = B + 2(1 + 2 + .... 277)
= 7161558 + 2.277(277 + 1) : 2
= 7238564
Câu 4 : \(\left(\frac{2^2}{2.4}+\frac{2^2}{4.6}+...+\frac{2^2}{34.36}\right)x-1\frac{1}{6}=1\frac{2}{3}\)
=> \(2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{34.36}\right)x-\frac{7}{6}=\frac{5}{3}\)
=> \(2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{34}-\frac{1}{36}\right)x=\frac{17}{6}\)
=> \(\left(\frac{1}{2}-\frac{1}{36}\right)x=\frac{17}{12}\)
=> x = 3
Câu 5 : Đặt A = 1 + 2 + 22 + ... + 29 (1)
=> 2A = 2 + 22 + 23 + ... + 210 (2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = (2 + 22 + 23 + ... + 210) - ( 1 + 2 + 22 + ... + 29)
A = 210 - 1 = 1024 - 1 = 1023
Câu 6 : Đặt A = 12 + 22 + 32 + .... + 1002
= 1.1 + 2.2 + 3.3 + ... + 100.100
= 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ... + 100(101 - 1)
= (1.2 + 2.3 + 3.4 + ... + 100.101) - (1 + 2 + 3 + 4 + ... + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 100.101.102 - 99.100.101
= 100.101.102
=> B = 343400
Khi đó A = B - (1 + 2 + 3 + 4 + ... + 100)
= 343 400 - [100.(100 + 1) : 2]
= 338 350