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\(\left(\frac{x}{20}+1\right)+\left(\frac{x-1}{21}+1\right)=\left(\frac{x-2}{22}+1\right)+\left(\frac{x-3}{23}+1\right)\)
\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)
\(\left(x+20\right).\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)
mà \(\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)\ne0\)
=> x+20=0 => x=-20
vậy x=-20
\(\frac{x}{20}+\frac{x-1}{21}=\frac{x-2}{22}+\frac{x-3}{23}\)
\(1+\frac{x}{20}+1+\frac{x-1}{21}=1+\frac{x-2}{22}+1+\frac{x-3}{23}\)
\(\frac{x+20}{20}+\frac{21+x-1}{21}=\frac{22+x-2}{22}+\frac{23+x-3}{23}\)
\(\frac{x+20}{20}+\frac{x+20}{21}=\frac{x+20}{22}+\frac{x+20}{23}\)
\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)
\(\left(x+20\right)\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)
Mà \(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\ne0\)
\(\Rightarrow x+20=0\)
\(\Rightarrow x=-20\)
Vậy x = -20
a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247x.(1233+20203333+303030333333+4242424233333333)=32
74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247x.(1233+2033+3033+4233)=32
74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247x.(3.433+4.533+5.633+6.733)=32
74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247x.33.(31−41+41−51+51−61+61−71)=32
74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247x.33.(31−71)=32
74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31+61+101+151+...+x.(x−1)2=20092007
26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62+122+202+302+...+(x−1).x2=20092007
22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32+3.42+4.52+5.62+...+(x−1).x2=20092007
2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−31+31−41+41−51+51−61+...+x−11−x1)=20092007
2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−x1)=20092007
1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2=20092007
2x=22009\frac{2}{x}=\frac{2}{2009}x2=20092
=> x = 2009
\(\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{3x+2y+z}{338}=\frac{169}{338}=\frac{1}{2}\)
\(\Rightarrow3x+25=\frac{1}{2}.144=72\)
\(\Leftrightarrow x=\frac{47}{3}\)
\(2y-169=\frac{1}{2}.25=\frac{25}{2}\)
\(\Leftrightarrow y=\frac{363}{4}\)
\(z+144=\frac{1}{2}.169=\frac{169}{2}\)
\(\Leftrightarrow z=\frac{-119}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{\left(3x+2y+z\right)+\left(25-169+144\right)}{144+25+169}=\frac{169+25-169+144}{144+25+169}=\)
\(\frac{1}{2}\)
Ta có
\(\frac{3x+25}{144}=\frac{1}{2}\Rightarrow6x+50=144\Rightarrow6x=94\Rightarrow x=\frac{47}{3}\)
\(\frac{2y-169}{25}=\frac{1}{2}\Rightarrow4y-338=25\Rightarrow4y=363\Rightarrow y=\frac{363}{4}\)
\(\frac{z+144}{169}=\frac{1}{2}\Rightarrow2z+288=169\Rightarrow2z=-119\Rightarrow z=\frac{-119}{2}\)
Hnay đi học, cô giáo có sửa cho bạn bài đó hong dọ, do cô mình giao cái bài về nhà y sì dãy í, mà mai nộp ròi, nhưng mình k biết làm, nếu bạn biết , chỉ mình với :((
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
ta có:
\(\frac{2x}{3}=\frac{3y}{4}=\frac{5z}{6}\)=>\(\frac{2x}{90}=\frac{3y}{120}=\frac{5z}{180}\)=>\(\frac{x}{45}=\frac{y}{40}=\frac{z}{36}\)
Áp dụng tính chất của dãy tỉ số băng nhau ta có:
\(\frac{x}{45}=\frac{y}{40}=\frac{z}{36}=\frac{x+y+z}{45+40+36}=\frac{121}{121}=1\)
=>\(\frac{x}{45}=1\)=>x=45
\(\frac{y}{40}=1\)=>y=40
\(\frac{z}{36}=1\)=>z=36
Vậy x=45;y=40;z=36
Bác viết nhộn đề gồi :v
\(.\frac{x+4}{20}+\frac{x+3}{21}+\frac{x+2}{22}+\frac{x+1}{23}=-4\)
\(\Rightarrow\frac{x+4}{20}+1+\frac{x+3}{21}+1+\frac{x+2}{22}+1+\frac{x+1}{23}+1=0\)
\(\Rightarrow\frac{x+24}{20}+\frac{x+24}{21}+\frac{x+24}{22}+\frac{x+24}{23}=0\)
\(\Rightarrow\left(x+24\right)\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)
=> x=-24
\(\frac{x+4}{20}+\frac{x+3}{21}\frac{x+2}{22}+\frac{x+1}{23}\)\(=-4\)
\(\Rightarrow\left(\frac{x+4}{20}+1\right)+\left(\frac{x+3}{21}+1\right)+\left(\frac{x+2}{22}+1\right)\)\(+\left(\frac{x+1}{23}+1\right)=0\)
\(\Rightarrow\left(\frac{x+4}{20}+\frac{20}{20}\right)+\left(\frac{x+3}{21}+\frac{21}{21}\right)\)\(+\left(\frac{x+2}{22}+\frac{22}{22}\right)+\left(\frac{x+1}{23}+\frac{23}{23}\right)=0\)
\(\frac{\Rightarrow x+24}{20}+\frac{x+24}{21}+\frac{x+24}{22}+\frac{x+24}{23}=0\)
\(\Rightarrow\left(x+24\right)+\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\ne0\)
\(\Rightarrow x+24=0\)
\(\Rightarrow x=24\)
Chúc bạn học tốt ( -_- )
Dài đấy :))
a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)
\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)
\(\Leftrightarrow\left|x-1\right|+8=9\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))
\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)
\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)
d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)
\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)
\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)
\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)
Vậy ta xét hai trường hợp sau :
1. \(x\ge-\frac{3}{16}\)
(*) <=>\(7x-2=4x+\frac{3}{4}\)
\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)
\(\Leftrightarrow3x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)
2. \(x< -\frac{3}{16}\)
(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)
\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)
\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)
\(\Leftrightarrow11x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)
Vậy x = 11/12
e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)
\(\Leftrightarrow x+1=4040\)
\(\Leftrightarrow x=4039\)
P/s đầu tiên mẫu số là 21
\(\frac{x+121}{21}+\frac{x+144}{22}+\frac{x+169}{23}=6\)
\(\left(\frac{x+121}{21}-1\right)+\left(\frac{x+144}{22}-2\right)+\left(\frac{x+169}{23}-3\right)=0\)
\(\frac{x+100}{21}+\frac{x+100}{22}+\frac{x+100}{23}=0\)
\(\left(x+100\right)\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}>0\)
Nên \(x+100=0\)
\(x=-100\)
Vậy \(x=-100\)