Bai 1:Tim x
a\(\frac{1}{9}=\frac{x}{27}\) b \(\frac{4}{x}=\frac{8}{6}\) c\(\frac{x}{3}-\frac{1}{2}=\frac{1}{5}\)
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Câu 6. Giải các phương trình sau:
a, x+frac{2x+frac{x-1}{5}}{3}1-frac{3x-frac{1-2x}{3}}{5}; b, frac{3x-1-frac{x-1}{2}}{3}-frac{2x+frac{1-2x}{3}}{2}frac{frac{3x-1}{2}}{5}-6
Câu 7. Giải các phương trình sau:
a, frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27}; b, left(frac{x+2}{98}+1right)+left(frac{x+3}{97}+1right)left(frac{x+4+++}{96}+1right)+left(frac{x+5}{95}+1right)
c, frac{x+1}{2004}+frac{x+2}{2003}frac{x+3}{2002}+frac{x+4}{2001}...
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Câu 6. Giải các phương trình sau:
a, x+\(\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\); b, \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}}{5}-6\)
Câu 7. Giải các phương trình sau:
a, \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\); b, \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4+++==}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
c, \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\); d, \(\frac{201-6}{99}+\frac{203-6}{97}=\frac{205-x}{95}+3=0\)
e, \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\); f, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
g, \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\); h, \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
i, \(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\);
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
a, tính:\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
b, tìm x biết:\(1\frac{1}{30}:(24\frac{1}{6}-24\frac{1}{5})-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=(-1\frac{1}{5}):(8\frac{1}{5}-8\frac{1}{3})\)
1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )
a)frac{-6}{11}.frac{5}{13}+frac{-6}{11}.frac{8}{13}-left(frac{-2}{5}right)^0 b)left(2frac{2}{3}+3frac{1}{2}right);left(4frac{3}{4}-2frac{1}{6}right)+frac{19}{31} c)2,4:left(-2right)^3+left(3-frac{9}{11}right).1frac{3}{8}
d)left(-frac{3}{4}right)^2:frac{-3}{8}+frac{1}{2}-frac{3}{4}-left(frac{-78}{57}right)^0
2. tìm x
a)x+frac{-1}{5}left(-frac{3}{4^{ }}right)^2 b)left|frac{5}{2}x+frac{2}{3}right|-frac{1}{4}0...
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1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )
\(a)\frac{-6}{11}.\frac{5}{13}+\frac{-6}{11}.\frac{8}{13}-\left(\frac{-2}{5}\right)^0\) \(b)\left(2\frac{2}{3}+3\frac{1}{2}\right);\left(4\frac{3}{4}-2\frac{1}{6}\right)+\frac{19}{31}\) \(c)2,4:\left(-2\right)^3+\left(3-\frac{9}{11}\right).1\frac{3}{8}\)
\(d)\left(-\frac{3}{4}\right)^2:\frac{-3}{8}+\frac{1}{2}-\frac{3}{4}-\left(\frac{-78}{57}\right)^0\)
2. tìm x
\(a)x+\frac{-1}{5}=\left(-\frac{3}{4^{ }}\right)^2\) \(b)\left|\frac{5}{2}x+\frac{2}{3}\right|-\frac{1}{4}=0\) \(c)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}+\frac{1}{2}x\) \(d)\left(x-\frac{1}{4}\right)^4=\frac{1}{81}\)
\(e)4x+3\frac{1}{4}=x-\frac{1}{4}\) \(g)\left(x-\frac{1}{3}\right)^3=\frac{1}{27}\)
Bai 1: a) \(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)
b) \(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
bài 2:
a) \(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)
b) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
c)
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
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Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
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Tìm x biết
a) x+2x+3x+4x+...+100x=-213
b)\(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
c)3(x-2)+2(x-1)=10
d)\(\frac{x+1}{3}=\frac{x-2}{4}\)
e)\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
f)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
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#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
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a) x + 2x + 3x + ... +100x = -213
=> x . (1 + 2 + 3 +... + 100) = - 213
=> x . 5050 = -213
=> x = - 213 : 5050
=> x = -213/5050
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)
=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)
=> \(x.\frac{1}{4}=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{1}{4}\)
=> \(x=\frac{2}{3}\)
c) 3(x-2) + 2(x-1) = 10
=> 3x - 6 + 2x - 2 = 10
=> 3x + 2x - 6 - 2 = 10
=> 5x - 8 = 10
=> 5x = 10 + 8
=> 5x = 18
=> x = 18:5
=> x = 3,6
d) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> \(4\left(x+1\right)=3\left(x-2\right)\)
=>\(4x+4=3x-6\)
=> \(4x-3x=-4-6\)
=> \(x=-10\)
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Bài 7: Giải các phương trình sau :
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
b) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
d) \(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
tim xa)2frac{3}{4}x-1frac{5}{8}x1b)2frac{3}{4}x-1frac{5}{8}x1c)frac{3}{2}x-frac{2}{5}frac{1}{3}x-frac{1}{4}d)2x-frac{1}{4}frac{5}{6}-frac{1}{2}xe)frac{-5}{6}+3xfrac{2}{3}-frac{1}{2}xf)frac{5}{2}x-frac{3}{2}x+frac{29}{10}g)frac{2}{3}+frac{7}{3}xfrac{5}{4}x+frac{1}{6}h)frac{1}{3}.x+frac{2}{5}left(x-1right)0giup mk nha moi nguoi
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tim x
a)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
b)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
c)\(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
d)\(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\)
e)\(\frac{-5}{6}+3x=\frac{2}{3}-\frac{1}{2}x\)
f)\(\frac{5}{2}x-\frac{3}{2}=x+\frac{29}{10}\)
g)\(\frac{2}{3}+\frac{7}{3}x=\frac{5}{4}x+\frac{1}{6}\)
h)\(\frac{1}{3}.x+\frac{2}{5}\left(x-1\right)=0\)
giup mk nha moi nguoi
Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
Xem thêm câu trả lời
tim x
a)\(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
b)\(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
giup mik vs
a) \(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
\(\left(-\frac{3}{4}+\frac{1}{6}\right).x=1-\frac{23}{9}\)
\(-\frac{7}{12}.x=-\frac{14}{9}\)
\(x=-\frac{14}{9}:\left(-\frac{7}{12}\right)\)
\(x=\frac{8}{3}\)
Vậy x = ...
b) \(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
\(\left|2x-\frac{3}{8}\right|+\frac{11}{4}=\frac{49}{16}\)
\(\left|2x-\frac{3}{8}\right|=\frac{49}{16}-\frac{11}{4}\)
\(\left|2x-\frac{3}{8}\right|=\frac{5}{16}\)
\(\Rightarrow\left|2x-\frac{3}{8}\right|\in\text{{}\frac{5}{16};-\frac{5}{16}\)}
Nếu, \(2x-\frac{3}{8}=\frac{5}{16}\)
\(2x=\frac{11}{16}\)
\(x=\frac{11}{32}\)
Nếu, \(2x-\frac{3}{8}=-\frac{5}{16}\)
\(2x=\frac{1}{16}\)
\(x=\frac{1}{32}\)
Vậy \(x\in\text{{}\frac{1}{32};\frac{11}{32}\)}
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