Cho biểu thức M=1/30 +1/42 +1/56 +1/72 +1/90 +1/110 +1/132
Chứng minh rằng M bé hơn 1/7
Tính giá trị biểu thức: 1/30 + 1/42 + 1/56+ 1/72+1/90+1/110+1/132
tính giá trị biểu thức ;A=1/30+1/42+1/56+1/72+1/90+1/110+1/132
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(\Rightarrow A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
\(A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}\)
\(A=1-\frac{1}{12}\)
\(A=\frac{11}{12}\)
tính giá trị biểu thức
A= 1/30+1/42+1/56+1/72+1/90+1/110+1/132
A=1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12
A=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/11-1/12
A=1/5-1/12
A=7/60
tính giá trị biểu thức
1/30+1/42+1/56+1/72+1/90+1/110+1/132
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}\)
Giá trị biểu thức là \(\frac{7}{60}\)
Mình chắc chắn
Tính giá trị biểu thức:
A = 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132
A=1/30 + 1/42 + ...+ 1/132
=1/5.6 + 1/6.7 +1/7.8 +....+1/11.12
=6-5/5.6 + 7-6/6.7+ ...+ 12-11/11.12
=1/5 -1/6 +1/6 - 1/7+....+ 1/11 - 1/12
=1/5-1/12
=7/60
A= 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10 + 1/10x11 + 1/11x12
A= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12
A= 1/5 - 1/12 = 7/60
A = 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132
A = 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12
A = 1/5 - 1/6 + 1/6 - 1/7 +......+1/11 - 1/12
A = 1/5 - 1/12
A =...........Tự mà tính nha!
Tính giá trị biểu thức:
A =1/30 + 1/42 + 1/56 + 1/72 + 1/90 +1/110 + 1/132
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}=\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{12-11}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
tính giá trị của biểu thức sau A=1/30+1/42+1/56+1/72+1/90+1/110+1/132
HELP ME!!!
\(A=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{5}-\dfrac{1}{12}\\ =\dfrac{12}{60}-\dfrac{5}{60}=\dfrac{7}{60}\)
a, cho hai phân số 1/n và 1/ n+1 n E Z và lớn hơn 0 chứng tỏ rằng tích của hai phân số bằng hiệu của chúng
b, áp dụng kết quả trên để tính giá tỉ biểu thức sau
A= 1/2*1/3+1/3*1/4+1/4*1/5+1/5*1/6+1/6*1/7+1/7*1/8+1/8*1/9
B=1/20+1/30+1/42+1/56+1/72+1/90+1/110
anh ê chơi thâm vừa thôi à nha
AK EM BẢO ANH NÈ EM NHỜ ANH CHỨ KO PHẢI EM TRẢ LỜI HỘ ANH
Tìm x khi:
1/3 - 1/12 - 1/20 - 1/30 - 1/42 - 1/56 - 1/72 - 1/90 - 1/110 = x - 5/13
Giải:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)
\(\Leftrightarrow x=\dfrac{-10}{143}\)
Vậy ...
Ta có:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}-\dfrac{1}{3}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\dfrac{8}{33}=x-\dfrac{28}{39}\\ x=-\dfrac{8}{33}+\dfrac{28}{39}\\ x=\dfrac{68}{143}\)
Vậy \(x=\dfrac{68}{143}\)