Hãy rút gọn biểu thức:
M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\)..........=\(\frac{10^2}{10.11}\)
A=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)
bn nào lm đc mk tick cx dễ đúng ko Vio vòng 2 thôi:D
A= (1x2x3x...x10)/(1x2x3x...x10)x(1x2x3x...x10)/(2x3x4x...x11)
A=1x 1/11=1/11
bạn nhớ nha
Hãy rút gọn biểu thức M=(12/1.2).(22/2.3).(32/3.4).....=102/10.11
Câu hỏi của pham nhu nguyen - Toán lớp 6 - Học toán với OnlineMath
\(\frac{1^2}{1.2}\) . \(\frac{2^2}{2.3}\) . \(\frac{3^2}{3.4}\) ..... \(\frac{10^2}{10.11}\)
\(\frac{1^2}{1.2}\)\(\frac{2^2}{2.3}\)\(\frac{3^2}{3.4}\).......\(\frac{10^2}{10.11}\)=..........
Đáp án : \(\frac{1}{11}\) các bạn giúp mình cách làm nha
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10}{11}\)
\(=\frac{1.2.3.....10}{2.3.4.....11}=\frac{1}{11}\)
Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^3}{3.4}.\frac{4^2}{4.5}\)
TÍNH GIÁ TRỊ BIỂU THỨC
hình như là 32 chứ k f 33
\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)
\(B=\frac{\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)}{\left(1\cdot2\right)\left(2\cdot3\right)\left(3\cdot4\right)\left(4\cdot5\right)}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot4\right)\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\left(2\cdot3\cdot4\cdot5\right)}\)
\(=\frac{1}{5}\)
\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)
\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)
\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1^2\cdot2^2\cdot3^2\cdot4^2\cdot5}=\frac{1}{5}\)
a) Phân tích đa thức thành nhân tử: x(x+2)(x2+2x+2)+1
b) Rút gọn biểu thức: A = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{[n\left(n+1\right)]^2}\)
Bạn thử giải câu này xem
NHỚ ĐỌC KỸ ĐỀ ĐẤY
https://olm.vn/hoi-dap/detail/211451950700.html?pos=476647086293
\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)
\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)
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:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Tính giá trị biểu thức:
\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
sorry mình nhầm
ta có:
M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\)
=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)
=\(\frac{1}{5}\)
vậy M=\(\frac{1}{5}\)
\(M=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)
ta có:
\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).
\(p=\frac{1.2.4+2.3.5+3.4.6+...+100.101.103}{1.2^2+2.3^2+3.4^2+...+100.101^2}\)Rút gọn P
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=\(\frac{A+B}{A-B}\)