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(x-13)/87+(x-27)/73+(x-67)/33+(x-73)/27=4
=>(x-13-87)/87+(x-27-73)/73+(x-67-33)/33+(x-73-27)/27=4-1-1-1-1
=>(x-100)/87+(x-100)/73+(x-100)/33+(x-100)/27=0
=>(x-100)*(1/87+1/73+1/33+1/27)=0
=>x-100=0
=>x=100
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
\(2x+\left(1+\dfrac{1}{6}\right)+\left(1+\dfrac{1}{12}\right)+\left(1+\dfrac{1}{20}\right)+\left(1+\dfrac{1}{30}\right)+\left(1+\dfrac{1}{42}\right)+\left(1+\dfrac{1}{56}\right)+\left(1+\dfrac{1}{72}\right)+\left(1+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
~ Chúc bạn học tốt ~
2x+\(\dfrac{7}{2.3}+\dfrac{13}{3.4}+\dfrac{21}{4.5}+...+\dfrac{91}{9.10}=10\)
\(2x+7\left(\dfrac{1}{2.3}\right)+13\left(\dfrac{1}{3.4}\right)+...+91\left(\dfrac{1}{9.10}\right)=10\)
\(2x+\left(7+13+...+91\right)\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)=10\)
\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
\(2x+336.\dfrac{2}{5}=10\)
\(2x+\dfrac{672}{5}=10\)
\(2x=10-\dfrac{672}{5}\)
\(2x=-\dfrac{622}{5}\)
\(x=-\dfrac{311}{5}\)
Tick nha 0o0^^^Nhi^^^0o0
\(4)\)
\(\dfrac{-\left(-x\right)}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{x}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\dfrac{2x}{10}-\dfrac{2}{10}=\dfrac{-10}{50}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{2x-2}{10}=\dfrac{-10-7}{50}\)
\(\dfrac{2x-2}{10}=\dfrac{-17}{50}\)
\(\Leftrightarrow50\left(2x-2\right)=-17.10\)
\(100x-100=-170\)
\(100x=-170+100=-70\)
\(x=-70:100=\dfrac{-7}{10}\)
\(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\left(x+1\right)\left(x-1\right)5.7\)
\(x\left(x-1\right)+1\left(x-1\right)=35\)
\(x^2-x+x-1=35\)
\(x^2-1=35\)
\(x^2=36\)
\(\Leftrightarrow x=\left\{\pm6\right\}\)
bạn có thể giải đc các bài còn lại k ? K phải mk ép bạn đâu nhưng nếu bạn lm đc thì giúp mk nha
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
a)\(12< 13;49>47\)
\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)
b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)
c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)
d)
\(1-\dfrac{67}{77}=\dfrac{10}{77}\)
\(1-\dfrac{73}{83}=\dfrac{10}{83}\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)
\(1-\dfrac{123}{128}=\dfrac{5}{128}\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)
b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)
\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)
c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)
d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)
\(\dfrac{73}{83}+\dfrac{10}{83}=1\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)
\(\dfrac{123}{128}+\dfrac{5}{128}=1\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)
\(\Rightarrow\left(\dfrac{x+24}{1996}+1\right)+\left(\dfrac{x+25}{1995}+1\right)+\left(\dfrac{x+26}{1994}+1\right)+\left(\dfrac{x+27}{1993}+1\right)+\left(\dfrac{x+2036}{4}-4\right)=0\)\(\Rightarrow\dfrac{x+2020}{1996}+\dfrac{x+2020}{1995}+\dfrac{x+2020}{1994}+\dfrac{x+2020}{1993}+\dfrac{x+2020}{4}=0\)\(\Rightarrow\left(x+2020\right)\left(\dfrac{1}{9996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)
\(\Rightarrow x+2020=0\Rightarrow x=-2020\)
\(\dfrac{x-13}{87}+\dfrac{x-27}{73}+\dfrac{x-67}{83}+\dfrac{x-73}{27}=4\)
<=>\(\dfrac{x-13}{87}-1+\dfrac{x-27}{73}-1+\dfrac{x-67}{83}-1+\dfrac{x-73}{27}-1=0\)
<=>\(\dfrac{x-100}{87}+\dfrac{x-100}{73}+\dfrac{x-100}{83}+\dfrac{x-100}{27}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{83}+\dfrac{1}{27}\right)=0\)
Do \(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{83}+\dfrac{1}{27}>0\)
=>x-100=0
<=>x=100
sửa chỗ :\(\dfrac{x-67}{33}\)