So sánh Q, P biết :
P = \(\dfrac{20}{30}\) + \(\dfrac{20}{70}\) + \(\dfrac{20}{126}\) + ... + \(\dfrac{20}{798}\)
Q = (\(\dfrac{31}{2}\) . \(\dfrac{32}{2}\) . \(\dfrac{33}{2}\) ... \(\dfrac{60}{2}\)) : ( 1.3.5...55)
Những câu hỏi liên quan
Tìm x :
a. \(x-\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-....-\dfrac{20}{53.55}\)
b. \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
Chứng minh \(⋮\) 31
giải các phương trình
a. \(|2-5x|=\left|3x+1\right|\)
b. \(\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
c. \(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)
\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)
\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
ĐXKĐ của phương trình \(:x\ne\pm5\)
\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)
Vậy phương trình vô nghiệm
\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)
Vậy \(x=-60\) là nghiệm của phương trình
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quy đồng
1. \(\dfrac{-5}{14},\dfrac{3}{20},\dfrac{9}{70}\)
2.\(\dfrac{10}{42},\dfrac{-3}{28},\dfrac{-55}{132}\)
3.\(\dfrac{7}{10},\dfrac{1}{33}\)
1:-5/14=-50/140
3/20=21/140
9/70=18/140
2: -55/132=-5/12=-35/84
10/42=20/84
-3/28=-9/84
3: 7/10=231/330
1/33=10/330
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Tính : A = \(\dfrac{\dfrac{1}{14}-\dfrac{1}{30}-\dfrac{1}{46}}{\dfrac{2}{35}-\dfrac{2}{75}-\dfrac{2}{115}}\):\(\frac{\frac{3}{8}-\frac{15}{17}+\frac{30}{31}}{\frac{1}{6}-\frac{20}{51}+\frac{40}{93}}\)
x-\(\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}=\dfrac{3}{11}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\dfrac{-2}{3}\left|x\right|+1\dfrac{1}{2}=\dfrac{2}{5}\)
các bạn phại đổi kiểu chữ để làm bài này (VNI) , (TELEX)
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x - \(\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-........-\dfrac{20}{53.55}=\dfrac{3}{11}\)
x - \(\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+\dfrac{20}{15.17}+...+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)
Đặt A = \(\dfrac{20}{11.13}+\dfrac{20}{13.15}+\dfrac{20}{15.17}+...+\dfrac{20}{53.55}\)
A = \(\dfrac{20}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+\dfrac{2}{15.17}+....+\dfrac{2}{53.55}\right)\)
A = \(\dfrac{20}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
A = \(10.\left(\dfrac{1}{11}-\dfrac{1}{55}\right)\)
A = 10 . 4/55
A = 8/11
TA Có : x - 8/11 = 3/11
x = 3/11 + 8/11
x = 11/11
x = 1
Vậy x = 1
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Xem thêm câu trả lời
1)Thực hiện phép tính
a) dfrac{9^8cdot5^6}{3^7cdot27^3cdot25^4} b) dfrac{5^{20}cdot6^{18}}{15^{20}cdot4^{10}} c) left(dfrac{1}{3}right)^{50}cdot9^{25}-left(dfrac{2}{3}right)^{100}:left(dfrac{8}{27}right)^{33} d) left(dfrac{3}{8}right)^{60}cdotleft(dfrac{2}{3}right)^{60}:left(dfrac{1}{2}right)^{119}cdotleft(dfrac{1}{2}right)
2) Tìm x, biết:
a) dfrac{x}{12}dfrac{5}{x} b)8^x2^{2x+3} c)dfrac{1}{2}sqrt{dfrac{1}{2}x-2}-dfrac{2}{3}dfrac{1}{3}
Đọc tiếp
1)Thực hiện phép tính
a) \(\dfrac{9^8\cdot5^6}{3^7\cdot27^3\cdot25^4}\) b) \(\dfrac{5^{20}\cdot6^{18}}{15^{20}\cdot4^{10}}\) c) \(\left(\dfrac{1}{3}\right)^{50}\cdot9^{25}-\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{8}{27}\right)^{33}\) d) \(\left(\dfrac{3}{8}\right)^{60}\cdot\left(\dfrac{2}{3}\right)^{60}:\left(\dfrac{1}{2}\right)^{119}\cdot\left(\dfrac{1}{2}\right)\)
2) Tìm x, biết:
a) \(\dfrac{x}{12}=\dfrac{5}{x}\) b)\(8^x=2^{2x+3}\) c)\(\dfrac{1}{2}\sqrt{\dfrac{1}{2}x-2}-\dfrac{2}{3}=\dfrac{1}{3}\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
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Giá trị biểu thức :
dfrac{1}{2}+left(dfrac{1}{2}right)^2+left(dfrac{1}{2}right)^3+left(dfrac{1}{2}right)^4+....+left(dfrac{1}{2}right)^{20} Là :
A dfrac{2^{20}-1}{2^{20}} B.dfrac{1}{2}-dfrac{1}{2^{20}}
C . 1 D . dfrac{2^{20}-2}{2^{20}}
- Ghi cách giải nữa nha!
Đọc tiếp
Giá trị biểu thức :
\(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+....+\left(\dfrac{1}{2}\right)^{20}\) Là :
A \(\dfrac{2^{20}-1}{2^{20}}\) B.\(\dfrac{1}{2}-\dfrac{1}{2^{20}}\)
C . 1 D . \(\dfrac{2^{20}-2}{2^{20}}\)
- Ghi cách giải nữa nha!
\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{20}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\right)\)
\(A=1-\dfrac{1}{2^{20}}=\dfrac{2^{20}-1}{2^{20}}\)
Chọn A
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Giải phương trình: \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}...\dfrac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)
\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)
\(\Rightarrow11\left(x-2\right)=-20x+40\)
\(\Rightarrow11x-22=-20x+40\)
\(\Rightarrow11x+20x=22+40\)
\(\Rightarrow31x=62\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
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a) dfrac{2}{1^2}.dfrac{6}{2^2}.dfrac{12}{3^2}.dfrac{20}{4^2}.dfrac{30}{5^2}.....dfrac{110}{10^2}.x-20
b) left(1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{2013}right).x+2013dfrac{2014}{1}+dfrac{2015}{2}+...+dfrac{4025}{2012}+dfrac{4026}{2013}
c) left(dfrac{1}{1.2}+dfrac{1}{3.4}+...+dfrac{1}{99.100}right).xdfrac{2012}{51}+dfrac{2012}{52}+...+dfrac{2012}{99}+dfrac{2012}{100}
Đọc tiếp
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)