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1) \(\frac{5-2n}{n-1}=\frac{5-2n+2-2}{n-1}=\frac{5-2-2.\left(n-1\right)}{n-1}=\frac{3}{n-1}-\frac{2.\left(n-1\right)}{n-1}=\frac{3}{n-1}+2\)
Để biểu thức trên nguyên thì \(\frac{3}{n-1}\) nguyên => \(3⋮n-1\)
=> \(n-1\inƯ\left(3\right)\)
=> \(n-1\in\left\{1;-1;3;-3\right\}\)
=> \(n\in\left\{2;0;4;-2\right\}\)
Vậy \(n\in\left\{2;0;4;-2\right\}\)
2) \(\frac{3n-4}{n-1}=\frac{3n-3-1}{n-1}=\frac{3.\left(n-1\right)-1}{n-1}=\frac{3.\left(n-1\right)}{m-1}-\frac{1}{n-1}=3-\frac{1}{n-1}\)
Để biểu thức trên nguyên thì \(\frac{1}{n-1}\) nguyên
=> \(1⋮n-1\)
=> \(n-1\inƯ\left(1\right)\)
=> \(n-1\in\left\{1;-1\right\}\)
=> \(n\in\left\{2;0\right\}\)
Vậy \(n\in\left\{2;0\right\}\)
c) \(\frac{6n-5}{2n-4}=\frac{6n-12+7}{2n-4}=\frac{3.\left(2n-4\right)+5}{2n-4}=\frac{3.\left(2n-4\right)}{2n-4}+\frac{5}{2n-4}=3+\frac{5}{2n-4}\)
Để biểu thức trên nguyên thì \(\frac{5}{2n-4}\) nguyên => \(5⋮2n-4\)
=> \(2n-4\inƯ\left(5\right)\)
Mà 2n - 4 là số chẵn \(\forall\) n nguyên nên không tìm được giá trị của n thỏa mãn vì 5 là số lẻ, không có ước chẵn
Vậy không tồn tại giá trị của n thỏa mãn đề bài
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)
\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)
Nên x + 116 = 0
<=> x = -116
a) \(x^2-2=\frac{1}{4}\)
\(\Rightarrow x^2=\frac{1}{4}+2\)
\(\Rightarrow x^2=\frac{9}{4}=2,25=1,5^2\)
\(\Rightarrow x=1,5\)
b) \(-\frac{3}{2}.\left(\frac{4}{5}+x\right)=1\frac{3}{2}\)
\(\Rightarrow-\frac{3}{2}.\left(\frac{4}{5}+x\right)=\frac{5}{2}\)
\(\Rightarrow\frac{4}{5}+x=\frac{5}{2}:-\frac{3}{2}\)
\(\Rightarrow\frac{4}{5}+x=-\frac{5}{3}\)
\(\Rightarrow x=-\frac{5}{3}-\frac{4}{5}\)
\(\Rightarrow x=-\frac{37}{15}\)
a. \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^4}=\frac{100^4}{100^5}=\frac{1}{100}\)
b. \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\frac{\left(-10\right)^5}{3^5}.\frac{\left(-6\right)^4}{5^4}=\frac{\left(-2.5\right)^5.\left(-2.3\right)^4}{3^5.5^4}=\frac{\left(-2\right)^5.5^5.\left(-2\right)^4.3^4}{3^5.5^4}=\frac{\left(-2\right)^9.5^5.3^4}{3^5.5^4}=\frac{\left(-2\right)^9.5}{3}=\frac{-512.5}{3}=\frac{-2560}{3}=\)
Ta có :
\(B=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)\)
\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x.\left(x+1\right)}{2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+...+\frac{x+1}{2}\)
\(B=\frac{2+3+4+...+\left(x+1\right)}{2}\)
để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)
\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)
\(\Rightarrow\)\(\left(x+3\right)x=23.20\)
\(\Rightarrow\)x = 20
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
1+3y/12=1+7y/4x=2+10y/12+4x=2(1+5y)/2(6+2x)
=1+5y/6+2x
do đó : 1+5y/6+2x=1+5y/5x<=>6+2x=5x<=>6=5x-2x
<=>3x=6=>x=2
Vậy x=2. chúc bạn học tốt
\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2010}}\)
\(5B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2009}}\)
Do đó 5B-B= \(\left(1+\frac{1}{5}+...+\frac{1}{5^{2009}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2010}}\right)\)
\(4B=1-\frac{1}{5^{2010}}\)
B = \(\frac{5^{2010}-1}{5^{2010}}:4\)
\(B=\frac{5^{2010}-1}{5^{2010}.4}\)
bạn có ghi sai đề ko vậy? phải là 5+1/5... ms đúng chứ?