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e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Rightarrow x=15\)
f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)
\(\Rightarrow x=2\)
a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)
Vậy \(x=1999\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)
Vậy \(x=45\)
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)
\(\Leftrightarrow x=45\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
Vậy x = 45.
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Bài cuối tương tự