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a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100
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a/ -7/2+3/4-17/12=-42/12+9/12-17/12=-50/12=-25/6
b/ 3/4+[-9/5]-4/3=45/60-108/60-80/60=-142/60=-71/30
c/ -1/12-[21/3-1/3]=-1/12-20/3=-1/12-80/12=-81/12=-27/4
d/ 22/7.[-7/11]+18= -2+18=16
e/ [3/4+2/3]:17/4-3/4=[9/12+8/12]:17/4-3/4=17/12:17/14-3/4=14/12-3/4=14/12-9/12=5/12
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Each term of S is n!(n2 + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!
By definition, n(n + 1)n! + n! = n! + n(n + 1)!
Therefore, S can be simplified as
1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!
So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)
\(=...\)
\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)
\(=\dfrac{101!+100\cdot101!}{101!}\)
\(=1+100=101\)
Hence, \(\dfrac{S+1}{101!}=101\)
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\(\frac{2}{5}+\left(\frac{3}{5}+\frac{3}{11}-\frac{3}{17}\right):\left(1+\frac{5}{11}-\frac{5}{17}\right)\)
\(=\frac{2}{5}+\frac{651}{935}:\frac{217}{187}\)
\(=\frac{2}{5}+\frac{3}{5}\)
\(=1.\)
c)
Chúc bạn học tốt!
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Bài 1:
b) Ta có: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}.\)
\(\Rightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{c^2}{16}.\)
\(\Rightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}\) và \(a^2-b^2+2c^2=108.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{a^2}{4}=4\Rightarrow a^2=16\Rightarrow\left[{}\begin{matrix}a=4\\a=-4\end{matrix}\right.\\\frac{b^2}{9}=4\Rightarrow b^2=36\Rightarrow\left[{}\begin{matrix}b=6\\b=-6\end{matrix}\right.\\\frac{c^2}{16}=4\Rightarrow c^2=64\Rightarrow\left[{}\begin{matrix}c=8\\c=-8\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(a;b;c\right)=\left(4;6;8\right),\left(-4;-6;-8\right).\)
Chúc bạn học tốt!
Mk sửa lại đề a
a) Tính (0,375-0,3+3/11+3/12)/(0,625-0,+5/11+5/12)