2/11.........3/11 <,>,= ?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Phương pháp giải:
Thực hiện phép trừ rồi điền kết quả vào chỗ trống.
Lời giải chi tiết:
11 − 5 = 6 11 − 7 = 4
11 − 8 = 3 11 − 2 = 9
11 − 6 = 5 11 − 4 = 7
11 − 9 = 2 11 − 3 = 8
a)
\(\left(\dfrac{3+2\sqrt{3}}{\sqrt{3}+2}-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =\left(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =3-2\\ =1\)
b)
\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)
a: \(=\left(\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{2+\sqrt{3}}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
=(căn 3-căn 2)(căn 3+căn 2)
=3-2=1
b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)
=(2-căn 11)(2+căn 11)
=4-11
=-7
Dãy này có: (10-1):1+1=10 số hạng
trung bình 1 số là: ( 10/11 +1/11):2=1/2
tổng của dãy là: 1/2*10=5 đ/s:5
\(\frac{1}{11}+\frac{2}{11}+\frac{3}{11}+\frac{4}{11}+...+\frac{10}{11}\)
\(=\frac{1+2+3+4+...+10}{11}\)
\(=\frac{\frac{10\times\left(10+1\right)}{2}}{11}\)
\(=\frac{55}{11}=5\)
\(\frac{1}{11}+\frac{2}{11}+\frac{3}{11}+\frac{4}{11}+...+\frac{10}{11}=\frac{1+2+3+4+...+10}{11}=\frac{55}{11}=5\)
Tính 11 – 2.
• Tách: 11 = 10 + 1.
• 10 – 2 = 8
• 8 + 1 = 9
• 11 – 2 = 9
Tính tương tự với các phép trừ còn lại ta có kết quả như sau:
11 – 2 = 9 11 – 3 = 8 11 – 4 = 7 11 – 5 = 6
11 – 6 = 5 11 – 7 = 4 11 – 8 = 3 11 – 9 = 2
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
11 – 4 = 7 11 – 5 = 6 |
11 – 8 = 3 11 – 6 = 5 |
11 – 9 = 2 11 – 2 = 9 |
11 – 3 = 8 11 – 7 = 4 |
a)3+3+3+3+3+3=3x6=18
b)5+5+5+5+5=5x5=25
c)11+11+11+11+11+11=11x6=66
2/11<3/11
2/11<3/11