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theo đề ra ta có Q=\(\left(\frac{51}{2}.\frac{52}{2}...\frac{100}{2}\right):\left(1.3....99\right)\)\(=\frac{1.2.3...50}{1.2.3...50}.\frac{51.52...100}{2.2.2....2.2}.\frac{1}{1.3....99}\)(50 thừa số 2)
\(=\frac{\left(1.2.3...50\right).\left(51.52...100\right).1}{\left(1.2.3...50\right).\left(2.2.2...2\right).\left(1.3...99\right)}\)\(=\frac{1.2.3.4....100}{\left(2.4.6.8..100\right).\left(1.3....99\right)}=\frac{1.2.3...100}{1.2.3...100}\)\(=1\)
các bạn thấy hay thì k cho mik nha
=(-1/2) : (-2/3) :( -3/4) :...: (-49/50)
= -1/2 . (-3/2) . (-4/3) . ... . (-50/49)
= -1/2.(-1/2) . (-50)
= - 1/100
\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
a) ĐK: \(x\ne0;x\ne-1\)
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{2}+x-2\right)\)
\(=\left(\frac{x+1-2+x}{\left(x^2+x\right)\left(x+1\right)}\right):\left(\frac{1+2x+4}{2}\right)\)
\(=\frac{2x-1}{\left(x^2+x\right)\left(x+1\right)}:\frac{2x+5}{2}\)\(=\frac{2\left(2x-1\right)}{\left(x^2+x\right)\left(x+1\right)\left(2x+5\right)}\)?? hình như hết tính tiếp được rồi :v
P/s: Có phải đề là tính giá trị biểu thức không?
= \(6.\frac{4}{6}-3.\frac{4}{6}-2.\frac{-2}{3}+4\) \(=4-2+\frac{4}{3}+4=6\frac{4}{3}\)
\(6.\left(-\frac{2}{3}\right)^2-3.\left(-\frac{2}{3}\right)^2-2:\left(-\frac{3}{2}\right)+4\)\(\)\(\)
\(=6.\frac{4}{9}-3.\frac{4}{9}-2.\left(-\frac{2}{3}\right)+4\)
\(=\frac{4}{9}.\left(6-3\right)-2.\frac{-2}{3}+4\)
\(=\frac{4}{9}.3-2.\left(-\frac{2}{3}\right)+4\)
\(=\frac{2}{3}.2-2.\left(-\frac{2}{3}\right)+4\)
\(=-\frac{2}{3}.\left(-2-2\right)+4\)
\(=-\frac{2}{3}.\left(-4\right)+4\)
\(=\frac{8}{3}+\frac{12}{3}\)
\(=\frac{16}{3}\)
f(x)=9x3-1/3x+3x2-3x+1/3x2-1/9x3-3x2-9x+27+3x
= 9x3-1/9x3+3x2+1/3x2-3x2-1/3-3x-9x+3x+27
= 80/9x3+1/3x2-28/3x+27
\(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):...:\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:...:\frac{-98}{99}:\frac{-99}{100}\)
\(=\frac{-1\cdot3\cdot4\cdot...\cdot99\cdot100}{2\cdot\left(-2\right)\cdot\left(-3\right)\cdot...\cdot\left(-98\right)\cdot\left(-99\right)}\)
\(=\frac{\left(-1\right)^{99}\cdot100}{2\cdot\left(-2\right)}=\frac{-1\cdot100}{-4}=\frac{-100}{4}=-25\)
- P/s: Không chắc chắn nhé!
