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\(T=1\cdot2\cdot3+2\cdot3\cdot4+...+20\cdot21\cdot22\)
\(4T=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+20\cdot21\cdot22\cdot4\)
\(4T=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+..+20\cdot21\cdot22\cdot\left(23-19\right)\)
\(4T=0\cdot1\cdot2\cdot3-1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+..+20\cdot21\cdot22\cdot23-19\cdot20\cdot21\cdot22\)
\(4T=21\cdot22\cdot23\cdot24\)
\(T=\frac{21\cdot22\cdot23\cdot24}{4}=21\cdot22\cdot23\cdot6=63756\\ k.cho.mk.nha\)
1.2.3+2.3.4+3.4.5+...+20.21.22
=1/4.(1.2.3.4+2.3.4.4+3.4.5.4+...+20.21.22.4)
=1/4.[1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+20.21.22.(23-19)]
=1/4.(1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+....-19.20.21.22+20.21.22.23)
=1/4.20.21.22.23
=53130
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
Xét \(M=x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1\)
\(\Rightarrow x^4M=x^{28}+x^{24}+x^{20}+x^{16}+...+x^8+x^4\)
\(\Rightarrow x^4M-M=\left(x^{28}+x^{24}+x^{20}+...+x^8+x^4\right)-\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)\)
\(\Rightarrow\left(x^4-1\right)M=x^{28}-1\)
\(\Rightarrow M=\dfrac{x^{28}-1}{x^4-1}\)
Xét \(N=x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1\)
\(\Rightarrow x^2N=x^{28}+x^{26}+x^{24}+x^{20}+...+x^4+x^2\)
\(\Rightarrow x^2N-N=\left(x^{28}+x^{26}+x^{24}+...+x^4+x^2\right)-\left(x^{26}+x^{24}+x^{22}+...+x^2+1_{ }\right)\)
\(\Rightarrow\left(x^2-1\right)N=x^{28}-1\)
\(\Rightarrow N=\dfrac{x^{28}-1}{x^2-1}\)
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
\(=\dfrac{M}{N}=\dfrac{\dfrac{x^{28}-1}{x^4-1}}{\dfrac{x^{28}-1}{x^2-1}}\)
\(=\dfrac{x^{28}-1}{x^4-1}.\dfrac{x^2-1}{x^{28}-1}=\dfrac{x^2-1}{x^4-1}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{1}{x^2+1}\)
Chúc bạn học tốt!
Câu 1 : \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}=\frac{x+2}{21}+\frac{x+2}{22}\)
=> \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}-\frac{x+2}{21}-\frac{x+2}{22}=0\)
=> x+2 . ( \(\frac{1}{18}+\frac{1}{19}+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\)) = 0
Vì \(\frac{1}{18}+\frac{1}{19}_{ }+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\ne0\)nên x+2=0
=> x= 0 - 2 = -2
Vậy x = -2
a: \(=\dfrac{2^{19}\cdot3^9+2^{20}\cdot3^{10}}{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}=\dfrac{2^{19}\cdot3^9\left(1+2\cdot3\right)}{2^{18}\cdot3^9\left(2+5\right)}=2\)
\(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+x^{18}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+x^{16}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
Rút gọn phân thức:
A=\(\frac{x^{24}+x^{20}+x^{16}+.....+x^4+1}{x^{26}+x^{24}+x^{22}+.......+x^2+1}\)
\(A=\frac{x^{24}+x^{20}+x^{16}+....+x^4+1}{x^{26}+x^{24}+x^{22}+.....+x^2+1}\) (1)
Ta có \(x^{26}+x^{24}+x^{22}+...+x^2+1\)
\(=\left(x^{26}+x^{22}+x^{18}+....+x^2\right)+\left(x^{24}+x^{20}+...+x^4+1\right)\)
\(=x^2\left(x^{24}+x^{20}+.....+x^4+1\right)+\left(x^{24}+x^{20}+...+x^4+1\right)\)
\(=\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+....+x^4+1\right)\) (2)
Từ (1),(2) ta có \(A=\frac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+....+x^4+1\right)}=\frac{1}{x^2+1}\)
Vậy A=\(\frac{1}{x^2+1}\)
Bài 3:
\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)
\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)
\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)
Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
Vậy x = 1
b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)
\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)
\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)
\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)
Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)
\(\Rightarrow x-1990=0\Rightarrow x=1990\)
a/ \(\frac{1}{3}.3^x+3^{x+2}=3^{16}+3^{13}\)
\(\Leftrightarrow3^{x-1}+3^{x+2}=3^{13}+3^{16}\)
\(\Leftrightarrow3^{x-1}\left(1+3^3\right)=3^{13}\left(1+3^3\right)\)
\(\Leftrightarrow3^{x-1}=3^{13}\Rightarrow x-1=13\Rightarrow x=14\)
b/ \(\frac{1}{6}6^x+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^{15}\left(1+6^3\right)\)
\(\Rightarrow x=16\)
c/ \(\frac{1}{2}2^{x+3}-2^x=2^{22}-2^{20}\)
\(\Leftrightarrow2^x\left(2^2-1\right)=2^{20}\left(2^2-1\right)\)
\(\Rightarrow x=20\)
Chọn D
Ta có x 3 + 1 x 20 + x - 1 x 2 22
Các số hạng có số mũ của x trùng nhau khi
với 
Mặt khác, 4k - 3 = 38
với 
Từ (2) => m lẻ
Suy ra trong khai triển trên có 4 số hạng có số mũ của x trùng nhau. Vậy sau khi khai triển và rút gọn T(x) có 21 + 23 - 4 =40 số hạng.