Mấy chế giúp pu pài này nhớ!^_^
Giải phương trình:
\(\frac{x-5}{1996}+\frac{x-15}{1986}=\frac{x-1986}{15}+\frac{x+1996}{5}\)
Thank mọi người nhju!!^_^
Giải PT\(\frac{x-29}{1986}+\frac{x-27}{1988}+\frac{x-25}{1990}+\frac{x-23}{1992}+\frac{x-21}{1994}+\frac{x-19}{1996}=\frac{x-1986}{29}+\frac{x-1988}{27}+\frac{x-1990}{25}+\frac{x-1992}{23}+\frac{x-1994}{21}+\frac{x-1996}{19}\)
pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015
Giải phương trình:
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
Mấy chế giúp mk pài này nhơ!!!^_^
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
\(\Rightarrow\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}+1\)
\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(\Rightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
Mà \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\)
\(\Rightarrow x+59=0\)
\(\Rightarrow x=-59\)
Chứng minh nghiệm phương trình sau là số nguyên:
\(\frac{x-1996}{15}+\frac{x-1975}{10}+\frac{x-1960}{5}+-2019=0\)
Bạn nào làm đúng mk sẽ t.i.c.k nha
ko vt lại đề
=> (x-1)/15 -133 + ( x-5)/10 -197 + x/5 -392 -2019 =0
=> (x-1)/15 + (x-5)/10 + x/5 - 2741=0
=> (2x-2)/30 + (3x-15)/30 + 6x/30 =2741
=> ( 2x-2+3x-15+6x)/30 =2741
=> 11x-17=82230
=> 11x= 82247
=> x= 7477
vì 7477 là số nguyên => nghiệm của phương trình là số nguyên
Vậy....
Giải phương trình
\(\frac{16}{\sqrt{x-1996}}+\frac{1}{\sqrt{x-1996}}=10-\left(\sqrt{x-1996}+\sqrt{y-2008}\right)\)
hướng dẫn thôi nhé
Có: \(\left(\frac{16}{\sqrt{x-1996}}+\sqrt{x-1996}\right)+\left(\frac{1}{\sqrt{y-2008}}+\sqrt{y-2008}\right)\)
\(\ge2\sqrt{\frac{16}{\sqrt{x-1996}}\sqrt{x-1996}}+2\sqrt{\frac{1}{\sqrt{y-2008}}\sqrt{y-2008}}=8+2=10\)
\(\Leftrightarrow\)\(\frac{16}{\sqrt{x-1996}}+\frac{1}{\sqrt{y-2008}}\ge10-\left(\sqrt{x-1996}+\sqrt{y-2008}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\frac{16}{\sqrt{x-1996}}=\sqrt{x-1996}\\\frac{1}{\sqrt{y-2008}}=\sqrt{y-2008}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2012\\y=2009\end{cases}}\)
giải phương trình sau:
a, \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
b,
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)
=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)
=> \(x-2004=0\)
=> \(x=2004\)
Vậy phương trình có nghiệm là x = 2004 .
b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)
=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình có nghiệm là x = 100 .
Giải phương trình sau
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}=0\)
Giải phương trình
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
Giải bất phương trình sau:
\(\frac{x+1994}{2009}+\frac{x+1995}{2010}>\frac{x+1996}{2011}+\frac{x+1997}{2012}\)
giải phương trình:
a/ \(\frac{x}{5}+\frac{2x+1}{3}=\frac{x-5}{15}\) b/\(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{16}{x^2-4}\)
giúp mình làm mấy câu này nha >.<