\(\frac{1}{9}^{2005}.9^{2005}-96^2:24^2\)
Những câu hỏi liên quan
\(\left(\frac{1}{9}\right)^{^{2005}}.9^{2005}-96^2:24^2\) = ?
\(\left(\dfrac{1}{9}\right)^{2005}.9^{2005}-96^2:24^2\)
\(\left(\dfrac{1}{9}\right)^{2005}.9^{2005}-96^2:24^2=\left(\left(\dfrac{1}{9}\right)^{2005}.9^{2005}\right)-\left(96^2:24^2\right)\)
\(=\left(\dfrac{1^{2005}}{9^{2005}}.9^{2005}\right)-\left(96^2:24^2\right)=\left(1^{2005}\right)-\left(\left(4.24\right)^2:24^2\right)\)
\(=1-\left(4^2\right)=1-16=-15\)
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Bài 2 :(1/9)^2005 . 9^2005 . 96^2 : 24^2
Bài 3 : Tìm các số a,b biết : a/2=b/3 và a+b = -15
tính hợp lí các biểu thức sau
\(\left(\frac{1}{9}\right)^{2005}.9^{2005}-96^2:24^2\)
\(16\frac{2}{7}:\left(\frac{-3}{5}\right)-28\frac{2}{7}:\left(\frac{-3}{5}\right)\)
\(\left(-2\right)^3.\left(\frac{3}{4}-0.25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
\(\left(\frac{1}{9}\right)^{2015}.9^{2015}-96^2:24^2=1^{2015}-4^2=1-16=-15\)
\(16\frac{2}{7}:\left(\frac{-3}{5}\right)-28\frac{2}{7}:\left(\frac{-3}{5}\right)=\left(16\frac{2}{7}-28\frac{2}{7}\right):\left(\frac{-3}{5}\right)=-12.\frac{-5}{3}=20\)
\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)=-8.\frac{1}{2}:\frac{13}{12}=-8.\frac{1}{2}.\frac{12}{13}=\frac{-48}{13}\)
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Tính giá trị của biểu thức:
a, 15/34 +7/21+19/34-20/15+3/7
b, 12-8×(3/2)^3
c,(1/9)^2005×9^2005-96^2÷24^2
\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(=>\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)
\(=>1+\frac{16}{21}-\frac{20}{15}\)
\(=>\frac{37}{21}-\frac{20}{15}\)
\(=>\frac{3}{7}\)
\(b,12-8\cdot\left(\frac{3}{2}\right)^3\)
\(=>12-8\cdot\frac{27}{8}\)
\(=>12-27\)
\(=>-15\)
\(c,\left(\frac{1}{9}\right)^{2005}\cdot9^{2005}-96^2:24^2\)
\(=>\left(\frac{1^{2005}^{ }}{9^{2005}}\cdot9^{2005}\right)-\left(96^2:24^2\right)\)
\(=>\left(1^{2005}\right)-16\)
\(=>1-16\)
\(=>-15\)
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bài 9:cho :
C=\(\frac{2005}{2}+\frac{20025}{3}+\frac{2005}{4}+...+\frac{2005}{2005}\)
D=\(\frac{2006}{1}+\frac{2007}{2}+\frac{2008}{3}+...+\frac{4004}{2004}\)
tính C-D
So sánh
a)A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)và B=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
b)M=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)và N=\(\frac{2009^{2009}-2}{2009^{2010}-2}\)
c)P=\(\frac{1+5+5^2+5^3+...+5^{10}}{1+5+5^2+5^3+...+5^9}\)và Q=\(\frac{1+3+3^2+3^3+...+3^{10}}{1+3+3^2+3^3+...+3^9}\)
a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)
\(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B Vay A<B
b,lam tuong tu nhu y a
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a)so sánh :920 và 2713 . Vì sao?
b) so sánh : A=\(\frac{2005^{2005}+1}{20005^{2006}+1}vàB=\frac{2005^{2004}+1}{2005^{2005}+1}\)vì sao?
a ) Ta có : \(9^{20}\)= \(\left(3^2\right)^{10}\)= \(3^{20}\)
\(27^{13}\)= \(\left(3^3\right)^{13}\)= \(3^{39}\)
Vì 39 > 20 => 9^ 20 < 27 ^ 13
Phần b bạn vào câu hỏi tương tự. Nhớ tích đúng cho tớ
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Tính nhanh
A=1×5×6+2×10×12+24×8×10/1×35+2×6×10+8×6×20
B=2006×2008-3/2005+2005×2008
C=18×123+9×4567×2+3×5310×6/1+4+7+.......+55+58-410
B=2006 * 2008 -3 / 2005 + 2005 * 2008
B=(2005+1)* 2008 -3 / 2005 +2005 *2008
B=2005 * 2008 + 2008 -3 / 2005 +2005*2008
B=2005 * 2008 + 2005 / 2005 +2005 * 2008
B= 1
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