\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)=?\)
Những câu hỏi liên quan
Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{^{x^2}}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
Thu gọn
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
\(B=\frac{\left(a+2008\right)!+\left(a+2009\right)!}{\left(a+2008\right)!-\left(a+2009!\right)}\)
\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)=?\)
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(-2)x\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)\)=?
a=\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)\)
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\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)=?\)
\(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)....\left(-1\frac{1}{2010}\right)=-2.\frac{-3}{2}.\frac{-4}{3}...\frac{-2011}{2010}=2.\frac{3}{2}.\frac{4}{3}...\frac{2011}{2010}\)
=2011
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Giải Phương trình
\(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x^2\right)-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)
\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)
Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)
2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)
\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)
\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
Vậy \(x=2003\)
3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)
\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)
\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)
Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)
\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)
Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)
\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)
Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)
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Tính:
\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)\)
\(=\left(-2\right).\left(-\frac{3}{2}\right).\left(-\frac{4}{3}\right)....\left(-\frac{2010}{2009}\right).\left(-\frac{2011}{2010}\right)=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2010\right).\left(-2011\right)}{2.3.4....2009.2010}=2011\)
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THU GỌN\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)