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d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)
\(-\frac{5}{6}\times x=\frac{5}{2}\)
\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)
\(x=-3\)
b.
\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)
\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)
\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=-\frac{19}{2}+\frac{37}{10}\)
\(3x=\frac{-95+37}{10}\)
\(3x=-\frac{58}{10}\)
\(3x=-\frac{29}{5}\)
\(x=-\frac{29}{5}\div3\)
\(x=-\frac{29}{5}\times\frac{1}{3}\)
\(x=-\frac{29}{15}\)
c.
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21-16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}\times\frac{4}{3}\)
\(x=\frac{5}{6}\)
d.
\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)
\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)
\(-\frac{2}{3}\times x=\frac{3-2}{10}\)
\(-\frac{2}{3}\times x=\frac{1}{10}\)
\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)
\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)
\(x=-\frac{3}{20}\)
e.
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{20+9}{12}\)
\(\left|x\right|=\frac{29}{12}\)
\(x=\pm\frac{29}{12}\)
Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)
f.
\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)
\(2x-\frac{1}{3}=\pm\frac{1}{6}\)
- \(2x-\frac{1}{3}=\frac{1}{6}\)
\(2x=\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{1+2}{6}\)
\(2x=\frac{3}{6}\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{2}\div2\)
\(x=\frac{1}{2}\times\frac{1}{2}\)
\(x=\frac{1}{4}\)
- \(2x-\frac{1}{3}=-\frac{1}{6}\)
\(2x=-\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{-1+2}{6}\)
\(2x=\frac{1}{6}\)
\(x=\frac{1}{6}\div2\)
\(x=\frac{1}{6}\times\frac{1}{2}\)
\(x=\frac{1}{12}\)
Vậy x = 1/4 hoặc x = 1/12.
Chúc bạn học tốt
Sorry nha, mik chép lộn đề
Làm lại câu a nha
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{12}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)
\(-\frac{5}{6}\times x=\frac{5}{12}\)
\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)
\(x=-\frac{1}{2}\)
Chúc bạn học tốt
a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)
⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)
Vậy: x=0
b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)
⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)
Vậy: x=-1
c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)
⇔\(2x=\frac{3\cdot6}{-1}=-18\)
hay x=-9
Vậy: x=-9
d) Ta có: \(\frac{x+1}{3}=\frac{3}{x+1}\)
⇔\(\left(x+1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: x∈{2;-4}
e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)
⇔\(9-x=\frac{-12\cdot5}{4}=-15\)
hay x=24
Vậy: x=24
f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)
⇔\(\left(x-1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy: x∈{5;-3}
g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)
⇔\(\left(5-x\right)^2=4\)
⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy: x∈{3;7}
h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)
⇔\(\left(4-x\right)^2=25\)
⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
Vậy: x∈{-1;9}
Cảm ơn bạn