Rút gọn: \(\frac{\frac{1}{77}-\frac{1}{777}-\frac{1}{7951}}{\frac{4}{77}-\frac{4}{777}-\frac{4}{4951}}\)
7+77+777+......+777...7 (n chữ số 7) =\(\frac{7}{81}\) (10 mũ n+1 -9*n-10)
\(\frac{3}{1x2}+\frac{3}{2x3}+\frac{3}{3x4}+\frac{3}{4x5}+\frac{3}{5x6}+....+\frac{3}{9x10}+\frac{77}{2x9}+\frac{77}{9x16}+\frac{77}{16x23}+...+\frac{77}{93x100}\)
\(\frac{4}{3x6}+\frac{4}{6x9}+\frac{4}{9x12}+\frac{4}{12x15}\) \(\frac{7}{1x5}+\frac{7}{5x9}+\frac{7}{9x13}+\frac{7}{13x17}+\frac{7}{17x21}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\) \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{138}+\frac{1}{340}\)
Toán quá dễ. Tự túc là hạnh phúc mọi nhà bn nhé !
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)
Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có :
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)
\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)
\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+\frac{7}{13.17}+\frac{7}{17.21}\)
\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{21}\right)\)
\(=\frac{7}{4}\cdot\frac{20}{21}=\frac{5}{3}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
Sủa đề : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)
Nêu định lý Pythagoras.
Rút gọn\(HÙNG=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{77\sqrt{78}+78\sqrt{77}}\)
Định lí Py-ta-go : Xét tam giác vuông có độ dài các cạnh góc vuông là a;b và cạnh huyền là c thì ta có
\(a^2+b^2=c^2\)
Và ngược lại , nếu có hệ thức trên thì tam giác đó cũng vuông
Bài kia :
Ta có tổng quát \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng ta được
\(H=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-....+\frac{1}{\sqrt{77}}-\frac{1}{\sqrt{78}}\)
\(=1-\frac{1}{\sqrt{78}}\)
\(3\frac{2}{5}+4\frac{3}{7}-\frac{1}{4}+\frac{44}{77}-2\frac{2}{5}-0,75\)
3 / 2 / 5 + 4 / 3 / 7 - 1 / 4 + 44 / 77 - 2 / 2 / 5 - 0,75
= 17 / 5 + 31 / 7 - 1 / 4 + 4 / 7 - 12 / 5 - 3 / 4
= ( 17 / 5 - 12 / 5 ) + ( 31 / 7 - 4 / 7 ) + ( 1 / 4 + 3 / 4 )
= 1 + 9 + 1
= 11
Bài làm
=17/5 + 31/7 - 1/4 +4/7 - 12/5 - 3/10
=( 17/5+12/5)-(31/7+4/7)-3/10
=29/5-5/1-3/10
=29/5-3/10-5/1
=61/10
Mình ko chác đâu nhé!
\(\frac{1}{2}+\frac{2}{8}+\frac{3}{28}+\frac{4}{77}+\frac{5}{176}\)
Rút gọn:
\(B=\frac{\sqrt{9-\sqrt{77}}+\sqrt{12.5-\sqrt{154}}}{\frac{1}{\sqrt{2}}-1}\)
bài 1 rút gọn
\(\frac{56}{24},\frac{33}{77}\)
NHANH LÊN NHÉ
\(\frac{56}{24}=\frac{7}{3};\frac{33}{77}=\frac{3}{7}\)
Tính \(D=\frac{2}{45}+\frac{1}{99}-\frac{2}{77}+\frac{4}{105}\)
D bằng [2/45+4/105]-[1/99+2/77]
26/315-25/693 bằng 23/495
Đ/S....................................................................
\(\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}{\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{2}.\frac{1}{3}.\frac{1}{4}}\)+\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}.\frac{1}{4}\)
Tính(rút gọn)