Chưng minh rằng :
\(\frac{1.2-1}{2\text{!}}+\frac{2.3-1}{3\text{!}}+\frac{3.\text{4}-1}{\text{4}\text{!}}+...+\frac{99.100-1}{100\text{!}}< 2\)
Những câu hỏi liên quan
tính tổng :F=\(\frac{\text{1+(1+2)+(1+2+3)+...+(1+2+3+...+100)}}{\text{1.2+2.3+...+99.100}}\)
chứng minh rằng
a)
\(\frac{1-2\text{s}in^2x}{2cot\left(\frac{\pi}{4}+\alpha\right).c\text{os}^2\left(\frac{\pi}{4}-\alpha\right)}=1\)
b)
\(\frac{\frac{\sqrt{3}}{2}c\text{os}2\text{a}-\frac{1}{2}sin2\text{a}}{1-\frac{1}{2}c\text{os}2\text{a}-\frac{\sqrt{3}}{2}sin2\text{a}}=tan\left(a+\frac{\pi}{4}\right)\)
CMR:
\(\frac{1}{1.2.3}+\frac{1}{2.3.\text{4}}+\frac{1}{3.\text{4}.5}+...+\frac{1}{98.99.100}=\frac{\text{4}9\text{4}9}{19800}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\left(\text{đpcm}\right)\)
\(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99.100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}=VP\) (đpcm)
Thực hiện phép tínha) frac{text{x + 9}}{x^2 - 9}-frac{text{3}}{text{x^2 + 3x}}b) frac{text{3x + 5
}}{text{x^2 - 5x
}}+frac{text{ 25 - x
}}{text{25 - 5x
}}c) frac{text{3
}}{text{2x
}}+frac{text{3x - 3
}}{text{2x - 1
}}+frac{ 2x^2 + 1
}{text{4x^2 - 2x
}}d) frac{text{1}}{text{3x - 2 }}-frac{1}{text{3x + 2 }}- frac{text{3x - 6}}{text{4 - 9x^2}}e) frac{text{18 }}{text{(x - 3)(x^2 - 9) }}-frac{text{3 }}{text{x^2 - 6x + 9 }}-frac{text{x}}{text{x^2 - 9}}g) frac{text{x + 2 }}{text{x + 3 }}-frac{text{...
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Thực hiện phép tính
a) \(\frac{\text{x + 9}}{x^2 - 9}-\frac{\text{3}}{\text{x^2 + 3x}}\)
b) \(\frac{\text{3x + 5 }}{\text{x^2 - 5x }}+\frac{\text{ 25 - x }}{\text{25 - 5x }}\)
c) \(\frac{\text{3 }}{\text{2x }}+\frac{\text{3x - 3 }}{\text{2x - 1 }}+\frac{ 2x^2 + 1 }{\text{4x^2 - 2x }}\)
d) \(\frac{\text{1}}{\text{3x - 2 }}-\frac{1}{\text{3x + 2 }}- \frac{\text{3x - 6}}{\text{4 - 9x^2}}\)
e) \(\frac{\text{18 }}{\text{(x - 3)(x^2 - 9) }}-\frac{\text{3 }}{\text{x^2 - 6x + 9 }}-\frac{\text{x}}{\text{x^2 - 9}}\)
g) \(\frac{\text{x + 2 }}{\text{x + 3 }}-\frac{\text{5 }}{\text{x^2 + x - 6 }}+\frac{\text{1}}{\text{2 - x}}\)
h) \(\frac{\text{4x }}{\text{x + 2 }}-\frac{\text{3x }}{\text{x - 2 }}+\frac{\text{12x}}{\text{x^2 - 4}}\)
i) \(\frac{\text{ x + 1 }}{\text{ x - 1 }}-\frac{\text{ x - 1 }}{\text{ x + 1 }}-\frac{\text{4}}{\text{1 - x^2}}\)
k) \(\frac{\text{
3x + 21
}}{\text{
x^2 - 9
}}+\frac{\text{2 }}{\text{x + 3 }}-\frac{\text{3}}{\text{x - 3}}\)
( Câu siêu khó dành cho học sinh đội tuyển Toán )( Giúp với )Bài 10 :b) frac{left(text{13}frac{text{1}}{text{4}}-text{2}frac{text{5}}{text{27}}-text{10}frac{text{5}}{text{6}}right).text{230}frac{text{1}}{text{25}}+text{46}frac{text{3}}{text{4}}}{left(text{1}frac{text{3}}{text{7}}+frac{text{10}}{text{3}}right):left(text{12}frac{text{1}}{text{3}}-text{14}frac{text{2}}{text{7}}right)}
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( Câu siêu khó dành cho học sinh đội tuyển Toán )
( Giúp với )
Bài 10 :
b) \(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}\)
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
Giải :
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
~~Học tốt~~
\(\text{Giải :}\)
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
\(\text{~~Học tốt~~}\)
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Bài 4: Tính hợp lý
A=\(\frac{4}{\text{1⋅2}}+\frac{4}{\text{3⋅5}}+......+\frac{4}{\text{20⋅11⋅2013}}\)
Bài 5: So sánh với 1:
A=\(\frac{1}{\text{1⋅2}}+\frac{1}{\text{2⋅3}}+\frac{1}{\text{3⋅4}}+......+\frac{1}{\text{49⋅50}}\)
Bài 5 :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{59}\)
\(A=1-\frac{1}{50}\)
từ trên ta có : \(1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)
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Ai giải giúp mấy bài toán vsBài 1:Asqrt{frac{1}{text{√}2+1}-frac{text{√}8-text{√}10}{2-text{√}5}}Bfrac{5text{√}5}{text{√}5+2}+frac{text{√}5}{text{√}5-1}-frac{3text{√}5}{3+text{√}5}Bài 2 rút gọn biểu thứcAleft(frac{x+sqrt[]{xy}}{text{√}x+text{√}y}-2right):frac{1}{text{√}x+2} với x :y 0Bleft(frac{a}{a-2text{√}a}+frac{a}{text{√}a-2}right):frac{text{√}a+1}{a-4text{√}a+4}Bài 3 cho biểu thứcPleft(frac{x-2}{x+2text{√}x}+frac{1}{text{√}x+2}right)frac{text{√}x+1}{text{√}x-1}a)Rút gọn Pb)tìm x để Ptext{√}...
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Ai giải giúp mấy bài toán vs
Bài 1:
A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)
B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)
Bài 2 rút gọn biểu thức
A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0
B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)
Bài 3 cho biểu thức
P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)
a)Rút gọn P
b)tìm x để P=\(\text{√}x+\frac{5}{2}\)
bài 4 rút gọn biểu thức
A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)
B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)
Bài 5
A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)
a)rút gọn A
b)tìm gtri x để A= -1/4
AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN
\(A=\frac{3}{\left(1\text{*}2\right)\text{*}\left(1\text{*}2\right)}+\frac{5}{\left(2\text{*}3\right)\text{*}\left(3\text{*}2\right)}+\frac{7}{\left(3\text{*}4\right)\text{*}\left(3\text{*}4\right)}+...............+\frac{19}{\left(9\text{*}10\right)\text{*}\left(10\text{*}9\right)}\)
theo bài ra ta có
n = 8a +7=31b +28
=> (n-7)/8 = a
b= (n-28)/31
a - 4b = (-n +679)/248 = (-n +183)/248 + 2
vì a ,4b nguyên nên a-4b nguyên => (-n +183)/248 nguyên
=> -n + 183 = 248d => n = 183 - 248d (vì n >0 => d<=0 và d nguyên )
=> n = 183 - 248d (với d là số nguyên <=0)
vì n có 3 chữ số lớn nhất => n<=999 => d>= -3 => d = -3
=> n = 927
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chứng tỏ rằng: A= \(\frac{\text{1}}{2\text{1}}\) +\(\frac{2}{3\text{1}}\) +...+ \(\frac{20\text{1}3}{20\text{1}4}\) < 1
sửa lại đề : Chứng tỏ rằng : A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)
bài làm
A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}\)
A = \(\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2014-1}{2014!}\)
A = \(1-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2014}{2014!}-\frac{1}{2014!}\)
A = \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2013!}-\frac{1}{2014!}\)
A = \(1-\frac{1}{2014!}< 1\)
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