Câu hỏi của pham nhu nguyen - Toán lớp 6 - Học toán với OnlineMath

\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)

\(M=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)

\(M=\frac{1.2.3.....10}{1.2.3....10}.\frac{1.2.3.....10}{2.3.4.....11}\)

\(M=1.\frac{1}{11}\)

\(M=\frac{1}{11}\)

Cảm ơn bạn nha

ta có :

\(a-b=1.2+\left(2.3-2^2\right)+\left(3.4-3^2\right)+..+\left(98.99-98^2\right)\)

\(=2+2+3+4+..+98\)

\(=1+\left(1+2+3+..+98\right)=1+98\times\frac{99}{2}=4852\)

A=1.2+2.3+3.4+4.5+5.6+...+2016.2017 

=> 3A = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+.......+2016.2017.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3) + .......+ 2016.2017.(2018-2015)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +..........+ 2016.2017.2018 - 2015.2016.2017

=> 3A = 2016.2017.2018

=> A = 2016.2017.2018 : 3 

Mày.ngu qua

a=1.2 + 2.3 +3.4+ ...+ 2010.2011\

3a=1.2.3+2.3.3+3.4.3+......+2010.2011.3

3a=1.2.3+2.3.(4-1)+3.4.(5-1)+............+2010.2011.(2012-2009)

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2010.2011.2012-2009.2010.2011

3a=2010.2011.2012

a=2010.2011.2012:3

a=?

A=2010.2011.2012:3

Bạn thử giải câu này xem

NHỚ ĐỌC KỸ ĐỀ ĐẤY

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\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)

\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)

Đặt: \(x^2+2x=t\)

khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)

\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)

b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)

Khi đó:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)

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

\(A=1-\frac{1}{\left(n+1\right)^2}\)

Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{3^{32}-1}{2}\)

Rút gọn: (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(316 + 1)(3³² + 1)

A=(3-1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3²-1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3⁴-1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3⁸-1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3¹⁶-1)(3¹⁶ + 1)(3³² + 1)
A=(3³² - 1)(3³² + 1)
A=3⁶⁴-1
=>A=(3⁶⁴-1) /2

Tử số = \(1.2.4+2.3.5+3.4.6+...+100.101.103\)

\(=1.2.\left(3+1\right)+2.3.\left(4+1\right)+3.4.\left(5+1\right)+...+100.101.\left(102+1\right)\)

\(=1.2.3+1.2+2.3.4+2.3+3.4.5+3.4+...+100.101.102+100.101\)

\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)+\left(1.2+2.3+3.4+...+100.101\right)\)

Mẫu số = \(1.2^2+2.3^2+3.4^2+...+100.101^2\)

\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+100.101.\left(102-1\right)\)

\(=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+100.101.102-100.101\)

\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)-\left(1.2+2.3+3.4+...+100.101\right)\)

đặt \(A=1.2.3+2.3.4+3.4.5+...+100.101.102\) và \(B=1.2+2.3+3.4+...+100.101\)

bạn tự tính : \(A=\frac{100.101.102.103}{4}=25.101.102.103\); \(B=\frac{100.101.102}{3}=100.101.34\)

rồi thay vào tìm P=\(\frac{A+B}{A-B}\)