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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 .
thanks
b, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1994}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1994}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)
=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)=0\)
=> \(x-2004=0\)
=> \(x=2004\)
Vậy phương trình có tập nghiệm là \(S=\left\{2004\right\}\)
a) Sửa đề: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
Ta có: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\)
nên x+36=0
hay x=-36
Vậy: x=-36
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow-21x=3x-60\)
\(\Leftrightarrow24x=60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{\left(8x-3\right)-2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)+\left(x+3\right)}{4}\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow x=-16\)
Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)
\(\Leftrightarrow27-9x+80-16x=12-12x-48\)
\(\Leftrightarrow-25x+107=-12x-36\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html
Mình cảm ơn trước nhaa
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)