Tính giá trị của biểu thức:
B = (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x ....... x (1 - 1/2003) x (1 - 1/2004)
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B= (1-1/2). ( 1-1/3).(1-1/4).(1-1/5)....(1-1/2004)
B= 1/2. 2/3 . 3/4. 4/5....2003/2004
B= 1/2004
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(B=\frac{1}{2004}\)
\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)
Rút gọn các thừa số ở tử và mẫu ta được:
\(B=\frac{1}{2004}\)
Đ/S:\(\frac{1}{2004}\)
Ta có:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)
Đơn giản hết sẽ là:
\(=\frac{1}{2004}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2004}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2003}{2004}\)
\(B=\dfrac{1\cdot2\cdot3\cdot...\cdot2003}{2\cdot3\cdot4\cdot...\cdot2004}\)
\(B=\dfrac{1}{2004}\)
B=(1-1/2)x(1-1/3)x(1-1/4)x(1-1/5)...(1-1/2003)x(1-1/2004)
B=1/2x2/3x3/4x4/5x...x2002/2003x2003/2004
B=1/2004
\(B=\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{2003}{2004}\)
\(B=\dfrac{1}{2004}\)
\(B=\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times....\left(1-\dfrac{1}{2003}\right)\times\left(1-\dfrac{1}{2004}\right)\)
\(B=\dfrac{1}{2}\times\dfrac{2}{3}\times....\times\dfrac{2002}{2003}\times\dfrac{2003}{2004}\)
\(B=\dfrac{1}{2004}\)
a) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right).\)
\(\Rightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(\Rightarrow A=\dfrac{1}{2004}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2004}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}...\dfrac{2003}{2004}\\ =\dfrac{1}{2004}\)
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdot....\cdot\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot....\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{2\cdot3\cdot4\cdot...\cdot2002\cdot2003}{3\cdot4\cdot5\cdot...\cdot2003\cdot2004}=\frac{1}{1002}\)