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Đề bài có lẽ sai thừa số thứ hai phải là 125125x127-127127x125
Khi đó thừa số thứ hai sẽ là
125x1001x127-127x1001x125=0
Tích trên sẽ bằng 0
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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
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(B=\frac{1}{2004}\)
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Ta có : \(\frac{0,36\cdot950+0,18\cdot726\cdot2+3\cdot324\cdot0,12}{1+3+5+7+9+...+27+29+31-152}\)
Đặt A = 0,36 . 950 + 0,18 . 726 . 2 + 3. 324 . 0,12
= 0,36 . 950 + 0,36 . 726 + 0,36 . 324
= 0,36(950 + 726 + 324) = 0,36.2000 = 720
Đặt B = (1 + 3 + 5 + 7 + 9 + ... + 27 + 29 + 31) - 152
Số số hạng : (31 - 1) : 2 + 1 = 16(số)
=> Tổng : (1 + 31).16 : 2 = 256
=> B = 256 - 152 = 104
Vậy \(\frac{A}{B}=\frac{720}{104}=\frac{90}{13}\)
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Bài 1:
\(S=1+3+5+7+...+297+299\)
Tổng trên là tổng các số hạng cách đều, số hạng sau hơn số hạng trước \(2\)đơn vị.
Số số hạng của tổng trên là: \(\left(299-1\right)\div2+1=150\)(số hạng)
Giá trị của tổng trên là: \(\left(299+1\right)\times150\div2=22500\)
Bài 2:
\(100-7\times\left(x-5\right)=58\)
\(\Leftrightarrow7\times\left(x-5\right)=100-58\)
\(\Leftrightarrow7\times\left(x-5\right)=42\)
\(\Leftrightarrow x-5=42\div7\)
\(\Leftrightarrow x-5=6\)
\(\Leftrightarrow x=6+5\)
\(\Leftrightarrow x=11\)
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\(=\frac{3}{7}+\frac{4}{9}\cdot\frac{2}{5}+\frac{5}{9}\cdot\frac{2}{5}-\frac{2}{5}\cdot1\)
\(=\frac{3}{7}+\frac{2}{5}\cdot\left(\frac{4}{9}+\frac{5}{9}-1\right)\)
\(=\frac{3}{7}+\frac{2}{5}\cdot0\)
\(=\frac{3}{7}+0\)
\(=\frac{3}{7}\)
( 1 + 3 + 5 + 7 +... + 2003 + 2005 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 125 x 1001 x 127 - 127 x 1001 x 125 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 0
= 0
~ Thiên Mã ~
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