Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 đềLàm 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) 2/3*x-3/2*x=5/12
x*(2/3-3/2)=5/12
x*(-5/6)=5/12
x=5/12:(-5/6)
x=-1/2
b)2/5+3/5*(3x-3,7)=-53/10
3/5*(3x-3,7)=(-53/10)-2/5=-57/10
(3x-3,7)=(-57/10):3/5=-19/2
3x=(-19/2)+3,7
3x=-29/5
x=(-29/5):3=-29/15
c) 7/9:(2+3/4*x)+5/9=23/27
7/9:(2+3/4*x)=8/27
2+3/4*x=21/8
3/4*x=5/8
x=5/6
k mình nha
a) 2/3.x-3/2.x=5/12
x.(2/3-3/2) =5/12
x.-5/6 =5/12
x =5/12:-5/6
x =1/2
b)2/5+3/5.(3x-3,7)=-53/10
3/5.(3x-3,7)=-53/10-2/5
3/5.(3x-3,7)=-57/10
3x-3,7 =-57/10:3/5
3x-3,7 =-19/2
3x =--19/2+3,7
3x =-29/5
x =-29/5:3
x =29/15
c)7/9:(2+3/4.x)+5/9=23/27
7/9:(2+3/4.x) =23/27-5/9
7/9:(2+3/4.x) =28/27
2+3/4.x =7/9:28/27
2+3/4.x =3/4
3/4.x =3/4-2
3/4.x =-5/4
x =-5/4:3/4
x =-5/3
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 \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
a) \(x.\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow x.\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow x=-\frac{1}{2}\)
b) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(\Leftrightarrow2+\frac{3}{4}x=\frac{21}{8}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{5}{8}\)
\(\Leftrightarrow x=\frac{5}{6}\)
c) \(\frac{3}{5}.\left(3x-3,7\right)=-\frac{57}{10}\)
\(\Leftrightarrow3x-3,7=-\frac{19}{2}\)
\(\Leftrightarrow3x=-\frac{29}{5}\)
\(\Leftrightarrow x=-\frac{29}{15}\)
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,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a. \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(\Rightarrow2+\frac{3}{4}x=\frac{7}{9}:\frac{8}{27}\)
\(\Rightarrow2+\frac{3}{4}x=\frac{21}{8}\)
\(\Rightarrow\frac{3}{4}x=\frac{21}{8}-2\)
\(\Rightarrow\frac{3}{4}x=\frac{5}{8}\)
\(\Rightarrow x=\frac{5}{8}:\frac{3}{4}\)
\(\Rightarrow x=\frac{5}{6}\)
b. \(\frac{-2}{3}x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow\frac{-2}{3}x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(\frac{-2}{3}\right)\)
\(\Rightarrow x=\frac{-3}{20}\)
c. \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Rightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Rightarrow\left|x\right|=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)
a, \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)
\(\frac{-5}{6}x=\frac{5}{12}\)
x = \(\frac{-1}{2}\)
Vậy x = \(\frac{-1}{2}\)
~~~
#Sunrise
a)2/3x-3/2x=5/12
x(2/3-3/2)=5/12
x(-5/6)=5/12
x=5/12/(-5/6)
x=-1/2