Tìm x \(\in\)N biết \(\frac{1}{3}+\frac{3}{35}<\frac{x}{210}<\frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)
1. Tìm x: \(\frac{x+1}{2005}+\frac{x+2}{2004}+\frac{x+3}{2003}+35=32\)
2.Cho A= 3+32+33+...+399
Tìm số tự nhiên n, biết rằng 2A+3=3n
Ta có 3A= \(^{3^2+3^3+3^4+...+3^{100}}\)
3A-A=2A= (\(3^2+3^3+3^4+...+3^{100}\))-(\(3+3^2+3^3+...+3^{99}\))
2A= \(3^{100}-3\)
theo bài ra ta có
2A+3=\(3^n\)= \(3^{100}-3+3=3^n\)=\(^{3^{100}}\)\(\Rightarrow\)n=100
Tìm \(x\in N\)
a, \(\frac{1}{3}+\frac{3}{35}
Tìm x, biết
1, \(x+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{32.35}=\frac{-29}{35}\)
Dat A=3/5.8 +3/8.11 +.........+3/32.35
A=1/5-1/8+1/8-1/11+1/11-1/14+.............+1/32-1/35
A=1/5-1/35
A=6/35
=>x+6/35=-29/35
=>x=-29/35-6/35
=>x=-1
Tìm \(x\in N\)biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)
\(-\frac{1}{x+1}=-\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(\Rightarrow x=2012\)
Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}:2\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{2003}{4006}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2003}{4006}-\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2002
Vậy x = 2002
Duyệt nha !!!
chúc hk tốt!!!
Bài 1:Tìm x ,biết:
a.\(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)
b.\(\frac{1}{x}=\frac{y}{2}-\frac{1}{3}\)
Bài 2:TÌm n\(\in\)N để phân số:B=\(\frac{10n-3}{4n-10}\)đạt GTLN.Tìm GTLN đó
a, \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)
\(\Rightarrow\frac{xy}{9y}-\frac{27}{9y}=\frac{1}{18}\Rightarrow y=2\)
\(\Rightarrow\frac{xy}{9y}-\frac{27}{9y}=\frac{1}{18}=\frac{2x}{18}-\frac{27}{18}=\frac{1}{18}\)
\(\Rightarrow2x-27=1\)
\(\Rightarrow2x=28\Rightarrow x=14\)
vậy x = 14
a, \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)
\(\Rightarrow\frac{xy}{9y}-\frac{27}{9y}=\frac{1}{9.2}\)
\(\Rightarrow9y=9.2\Rightarrow y=2\)
thay y = 2 vào ta có :
\(\frac{2x}{18}-\frac{27}{18}=\frac{1}{18}\)
\(\Rightarrow2x-27=1\Rightarrow2x=28\Rightarrow x=14\)
b, \(\frac{1}{x}=\frac{y}{2}-\frac{1}{3}\)
\(\Rightarrow\frac{1}{x}=\frac{3y}{6}-\frac{2}{6}\)
\(\Rightarrow\frac{1}{x}=\frac{3y-2}{6}\)
\(\Rightarrow x=6\)
2. \(B=\frac{10n-3}{4n-10}=\frac{\frac{5}{2}.\left(4n-10\right)+22}{4n-10}=\frac{5}{2}+\frac{22}{4n-10}\)
để \(B\) có giá trị lớn nhất thì \(\frac{22}{4n-10}\) là số dương lớn nhất
=> 4n - 10 là số dương nhỏ nhất ( n thuộc N )
\(\Rightarrow4n-10=2\Rightarrow4n=12\Rightarrow n=3\)
ta có :
\(B=\frac{10n-3}{4n-10}=\frac{30-3}{12-10}=\frac{27}{2}\)
Vậy để \(B\) có giá trị lớn nhất thì \(n=3\)
giá trị lớn nhất của \(B=\frac{27}{2}\)
Bài 1: Tìm x
\(\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)
\(\frac{3}{10}-\frac{3}{2x+4}=\frac{24}{35}\)
\(\frac{3}{2x+4}=\frac{-27}{70}\)
tự làm nốt
\(\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+....+\frac{3}{x\left(x+2\right)}=\frac{24}{35}\)
\(\Leftrightarrow3\left(\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{x\left(x+2\right)}\right)=\frac{24}{25}\)
\(\Leftrightarrow3\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{24}{35}\)
\(\Leftrightarrow3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{24}{35}\)
\(\Leftrightarrow3\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)
\(\Leftrightarrow\frac{3}{5}-\frac{3}{x+2}=\frac{24}{35}\)
\(\Leftrightarrow\frac{3}{x+2}=\frac{3}{5}-\frac{24}{35}\)
\(\Leftrightarrow\frac{3}{x+2}=\frac{21-24}{35}\)
\(\Leftrightarrow\frac{3}{x+2}=\frac{3}{-35}\)
\(\Rightarrow x+2=-35\Leftrightarrow x=-35-2\Leftrightarrow x=-37\)
Vậy \(x=-37\)
Tìm x biết : \(|2x-1|-3=\left(26+\frac{4}{2013}\right).\left(\frac{-34}{35}+\frac{2}{5}-\frac{3}{7}+1\right)\)
Giải :|2x−1|- 3 = 0
|2x−1| = -3
2x+1= -3
2x= -3 -1
2x=-4
x=-2 hoặc x= 2 (Vì loại bỏ số hạng chứa giá trị tuyệt đối. Điều này tạo ra 1 ± ở vế phải của phương trình vì |x|=±x)
\(\Rightarrow\)x= ±2
Tìm số nguyên x biết: (\(\left(\frac{-5}{3}\right)^3
Tìm số nguyên x biết : \(\left(\frac{-5}{3}\right)^3
\(\left(-\frac{5}{3}\right)^3=\left(-\frac{5}{3}\right).\left(-\frac{5}{3}\right).\left(-\frac{5}{3}\right)=-\frac{125}{27}\)
Đáp số: \(-\frac{125}{27}\)
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