\(\frac{\left(2^3.5^4.11\right).\left(2.5^3.11^2\right)}{\left(2^2.5^311\right)^2}\)
Những câu hỏi liên quan
\(\frac{\left(2^3.5^4.11\right).\left(2.5^3.11^2\right)}{\left(2^2.5^311\right)^2}\)
làm ơn hãy trả lời đi mình sẽ cho 1000 điểm
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\(B=\dfrac{\left(2^3.5^4.11\right).\left(2.5^2.11^2\right)}{\left(2^2.5^3.11\right)^2}\)
\(B=\dfrac{\left(2^3.5^4.11\right).\left(2.5^2.11^2\right)}{\left(2^2.5^3.11\right)^2}\)
\(\Leftrightarrow B=\dfrac{2^3.5^4.11.2.5^2.11^2}{2^4.5^9.11^2}\)
\(\Leftrightarrow B=\dfrac{2^4.5^6.11^3}{2^4.5^9.11^2}\)
\(\Leftrightarrow B=\dfrac{11}{5^3}\)
\(\Leftrightarrow B=\dfrac{11}{125}\)
Vậy...
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\(B=\dfrac{2^4\cdot5^6\cdot11^3}{2^4\cdot5^6\cdot11^2}=11\)
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= \(\dfrac{\left(2^3.5^4.11\right).\left(2.5^2.11^2\right)}{\left(2^2.5^3.11\right)^2}\)
= \(\dfrac{2^4.5^6.11^3}{\left(2^2.5^3.11\right).\left(2^2.5^3.11\right)}\)
= \(\dfrac{2^4.5^6.11^3}{2^4.5^6.11^2}\)
= 11
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Thực hiện phép tính:
a) \(\frac{\left(2^3.3^3\right)^2.5^5}{\left(2^2\right).\left(3^2.5^2\right)^2}\)
b) \(\frac{2^3.11-2^3.8}{6.\left(-1\right)^{2014}}\)
C = ( 2^3.5^4.11).(2.5^3.11^2)/(2^2.5^3.11)^2
\(C=\frac{\left(2^3.5^4.11\right).\left(2.5^3.11^2\right)}{\left(2^2.5^3.11\right)^2}\)
\(C=\frac{2^4.5^7.11^3}{2^4.5^6.11^2}\)
\(C=5.11\)
\(C=55\)
Chúc bn học tốt !!!!
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R = \(\left\{2015-2016^0.\left[2^3.5-\left(-1\right)^{2016}.\frac{1}{2^{19}}.\left(2.5^2-2^4.3\right)^{20}\right]\right\}-10^3\)
THỰC HIỆN PHÉP TÍNH:
\(\frac{\left(2^3.3^3\right)^2.5^5}{\left(2^2\right)^3.\left(3^2.5^2\right)^2}\)
Rút gọn biểu thức :
B=\(\left(\frac{3}{5}\right)^2.5^2-\left(2\frac{1}{4}\right)^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=\left(\frac{3}{5}\right)^2\cdot5^2-\left(2\frac{1}{4}\right)^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=\left(\frac{3}{5}\cdot5\right)^2-\left(\frac{9}{4}:\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=3^2-\left(\frac{9}{4}\cdot\frac{4}{3}\right)^3+\frac{1}{2}\)
\(B=3^2-3^3+\frac{1}{2}=-18+\frac{1}{2}=-\frac{35}{2}\)
Rút gọn phân số:
a)\(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}\)
b)\(\frac{2^3.3^4}{2^2.3^2.5};\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}\)
c)\(\frac{121.75.130.169}{39.60.11.198}\)
d)\(\frac{1998.1990+3978}{1992.1991-3984}\)
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
b) \(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{2\cdot9}{5}=\frac{18}{5}\)
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot1\cdot11\cdot1}{1\cdot5\cdot7\cdot1}=\frac{22}{35}\)
c) \(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11\cdot11\cdot13\cdot10\cdot169}{13\cdot3\cdot6\cdot10\cdot11\cdot11\cdot6\cdot3}\)
\(=\frac{169}{3\cdot6\cdot6\cdot3}=\frac{169}{324}\)
d) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)
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chứng tỏ rằng với mọi n thuộc N* ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)
\(=\frac{n}{2\left(3n+2\right)}\)
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