Tính:
a, A= \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}+6^{11}}\)
b, B= \(\frac{1}{\frac{1-1}{1-2^{-1}}}\) + \(\frac{1}{1+\frac{1}{1+2^{-1}}}\)
Những câu hỏi liên quan
1) Tinh gia tri cua bieu thuc:
A=\(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
B=\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)
\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0
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Câu 1: Tính
a) A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{98^2}-1\right).\left(\frac{1}{99^2}-1\right)\)
b) B=\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:\left(-1\frac{1}{100}\right)\)
c) C=\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^4}\)
Câu 1:(2^-1+3^-1):(2^-1-3^-1)+(2^-1.2^0):2^3
Câu 2: \(\left(-\frac{1}{3}\right)-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
Câu 3: \(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
Câu 4: \(\frac{1}{1-\frac{1}{1-2^{-1}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}\)
Câu 1: \(\frac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}+\frac{2^{-1}.1}{2^3}=\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{2}-\frac{1}{3}}+\frac{\frac{1}{2}}{8}=\frac{\frac{5}{6}}{\frac{1}{6}}+\frac{1}{6}=\frac{30}{6}+\frac{1}{6}=\frac{81}{16}\)
Câu 2:\(\frac{-1}{3}-1+\frac{\frac{1}{4}}{2}=\frac{-4}{3}+\frac{1}{8}=\frac{-29}{24}\)
Câu 3:\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{10}\left(2.3^2-3\right)}=\frac{2^{11}.3^{10}\left(2+2.5\right)}{2^{11}.3^{10}\left(2.3^2-3\right)}=\frac{4}{5}\)
Câu 4: \(\frac{1}{1-\frac{1}{1-2^{-1}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}=\frac{1}{1-\frac{1}{1-\frac{1}{2}}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}=\frac{1}{1-\frac{1}{\frac{1}{2}}}+\frac{1}{1+\frac{1}{\frac{3}{2}}}=-1+\frac{3}{5}=\frac{-2}{5}\)
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#Đoàn Thị Huyền Đan ơi: Câu 1 với câu 4 thì đúng rồi còn câu 2 với 3 thì sai k/q rồi nhé!
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\(^{\left(\frac{5^4-5^3}{125^4}\right)^3}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\frac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\frac{1}{3}\right)^3.\left(1\frac{1}{2}\right)^4.\left(-1\right)^5}\)
Câu 1: Tính
a) A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{98^2}-1\right).\left(\frac{1}{99^2}-1\right)\)
b) B=\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:\left(-1\frac{1}{100}\right)\)
c) C=\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^4}\)
Tính: a. A=\(\frac{4^6.9^5+6^9+120}{-8^4.3^{12}-6^{11}}\)
b. B=\(\frac{1}{1-\frac{1}{1-2^{-1}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}\)
Tính B=\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{5}{12}\right)^2}-\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
giúp mình giải chi tiết nha các bạn mai mình phải nộp rồi
qwertyuiopasdfgggggghjkllzxcvbnmm,.//234567890-=`
Chịu khó đọc lại đi dễ mà
1, tìm x
\(a.\frac{-3}{4}-y:\frac{1}{5}=\frac{9}{28}\)
\(b.5^x+5^{x+2}=650\)
2.
tính
\(M=\frac{4^6.9^3+6^9.120}{-8^4.3^{12}-6^{11}}\)
help me ! nhờ các bạn giúp mk với
Bài 1
a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)
y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)
y= \(\frac{-3}{14}\)
b.5x + 5x+2=650
5x . 1 + 5x + 52=650
5x(1+25)=650
5x.26=650
5x=25
x=2
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Bài: Rút gọn :
a) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}\)
b) \(4^6.256^2.2^4\)
c) \(A=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
d) \(B=\frac{4^2.25^2+32.125}{2^3.5^2}\)
a) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{1.81^2}{243}.\frac{3^2}{3^2}=\frac{6561}{243}.1=27\)
b, \(4^6.256^2.2^4=2^{12}.2^{16}.2^4=2^{32}\)
c) \(A=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)(Mình rút gọn lun cho nhanh nhé ) \(\Rightarrow A=\frac{4}{5}\)
d) \(\Rightarrow B=70\)k cho mình nha Cô Nàng Họ Dương
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Đây nhé : ý a,b mình đã giải thích rồi
c) \(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{12}{15}=\frac{4}{5}\)\(\frac{4}{5}\)
d) \(=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}=\frac{2^4.5^3.\left(5+2\right)}{2^3.5^2}=2.5.7=70\)
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