1, tính nhanh:
(1-1/2) *(1-1/3) *(1-1/4)* (1-1/5)*......*(1-1/2003)*(1-1/2004)
tính nhanh ạ (1-1/2)x(1-1/3)x(1-1/4)x(1-1/5).....(1-1/2003).(1-1/2004)
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(A=\dfrac{1}{2004}\)
Tính nhanh;
( 1- 1/2 ) x ( 1 - 1/3 ) x ( 1- 1/4 ) x ( 1 - 1/5 ) x......x ( 1 - 1 / 2003 ) x ( 1 - 1/ 2004 )
Tìm x
1 / 2 x X - 3 /4 = 5/6
1) =1/2 x 2/3 x 3/4 x 4/5 x .... x 2002/2003 x 2003/2004
=1/2004
2) 1/2 x X-3/4=5/6
1/2 x X =3/4+5/6
1/2 x X =19/12
X=19/6
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)
\(\frac{1}{2}.x-\frac{3}{4}=\frac{5}{6}\)
\(\frac{1}{2}.x=\frac{5}{6}+\frac{3}{4}\)
\(\frac{1}{2}.x=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
\(x=\frac{19}{12}:\frac{1}{2}\)
\(x=\frac{19}{12}.2=\frac{19}{6}\)
(1-1/2) x (1-1/3) x (1-1/4) x...x (1-1/2004)
=1/2 x 2/3 x 3/4 x ... x 2002/2003 x 2003/2004 = 1/2004
1/2 x X -3/4 = 5/6
=> 1/2 x X = 5/6 +3/4 = 19/12
=> x= 19/12 : 1/2
=> x=19/6
tính nhanh
(1-1/2)x(1-1/3)x(1-1/4)x......(1-1/2003)x(1-1/2004)
\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}=\frac{1x2x3x...x2002x2003}{2x3x4x...x2003x2004}=\frac{1}{2004}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)
Tính nhanh
(1 - 1/ 2) x (1 - 1/3) x ( 1 - 1/4 ) x ... x (1 - 1/2003) × ( 1 - 1/ 2004 )
ta sẽ ra được kết quả qua cách giảm ước của cả tử và mẫu . vậy cuối cùng nhìn lại trên tử còn 1 mẫu thì còn 2004 vậy phân số ra được là 1/2004
Tính nhanh :
( 1 - 1/2 ) × ( 1 - 1/3 ) × ( 1 - 1/4 ) × ... × ( 1 - 1/2003 ) × ( 1 - 1/2004 )
= 1/2 × 2/3 × 3/4 × ... × 2002/2003 × 2003/2004
= 1 × 2 × 3 × ... × 2002 × 2003 / 2 × 3 × 4 × ... × 2003 × 2004
= 1/2004
(1 - 1/ 2) x (1 - 1/3) x ( 1 - 1/4 ) x ... x (1 - 1/2003) × ( 1 - 1/ 2004 )
Giải
( 1 - 1/2 ) × ( 1 - 1/3 ) × ( 1 - 1/4 ) × ... × ( 1 - 1/2003 ) × ( 1 - 1/2004 )
= 1/2 × 2/3 × 3/4 × ... × 2002/2003 × 2003/2004
= 1 × 2 × 3 × ... × 2002 × 2003 / 2 × 3 × 4 × ... × 2003 × 2004
= 1/2004
tính nhanh nha:
B = ( 1 - 1/2 ) x ( 1 - 1/3 ) x ( 1 - 1/4 ) x ( 1 - 1/5 ) x ....x ( 1 - 1/2003 ) x ( 1 - 1/2004 )
ai đó giải hộ mình. Minhd đang cần gấp lắm mính tick nhanh cho
\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)
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 = \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x......x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)
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 = \(\frac{1x2x3x4x..........x2002x2003}{2x3x4x5x.....x2003x2004}\)
Sau khi trực tiêu các số thì ta có:
B = \(\frac{1}{2004}\)
bài 1 : (4đ) 1) Tính : A = 1 phần 2003 + 1 phần 2004 - 1 phần 2005 : 5 phần 2003 + 5 phần 2004 - 5 phần 2005 - ( qua phân số khác rồi nhé ) 2/2002 + 2/2003 - 2/2004 : 3/2002 + 3/2003 - 3/2004 2) Cho B = 1/3+1/3 mũ 2 + 1/3 mũ 3 + 1/3 mũ 4 + ... +1/3 mũ 2015 + 1/3 mũ 2016 . Chứng minh ràng B<1/2
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tính :
( 1-1/2 ) * ( 1-1/3 ) * ( 1-1/4 ) * ( 1-1/5 ) * ..... * ( 1 -1/2003 ) * ( 1-1/2004 )
Đề bài
= \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2002}{2003}\times\frac{2003}{2004}\)
= \(1\times2\times3\times4\times...\times2002\times2003/2\times3\times4\times5\times...2003\times2004\)
= \(\frac{1}{2004}\)
Đề bài
= 1/2 x 2/3 x 3/4 x 4/5 x .... x 2002/2003 x 2003/2004
= 1 x 2 x 3 x 4 x ...x 2002 x 2003 / 2 x 3 x 4 x 5 x .... x 2003 x 2004
= 1/2004
K nha
tinh nhanh. A=(1-1/2)*(1-1/3)*(1-1/4) ... *(1-1/2003)*(1-1/2004)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}=\frac{1}{2004}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2004}\right)\)
\(\Rightarrow A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}...\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(\Rightarrow A=\frac{1\cdot2\cdot3...2003}{2\cdot3\cdot4...2004}=\frac{1}{2004}\)