Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
g)=>x+1/2=0
x=0-1/2
x=-1/2
hoặc 2/3-2x=0
2x=2/3-0
2x=2/3
x=2/3:2
x=1/3
nhìn @_@ hoa cả mắt đăng từng bài thôi bạn
a) \(x+\frac{3}{4}=\frac{-3}{7}\Leftrightarrow x=\frac{-3}{7}-\frac{3}{4}=-\frac{33}{28}\)
b) \(\frac{5}{2}-\left(\frac{3}{2}+x\right)=\frac{7}{4}+\frac{3}{5}=\frac{47}{20}\Rightarrow\frac{3}{2}+x=\frac{5}{2}-\frac{47}{20}=\frac{3}{20}\Rightarrow x=\frac{3}{20}-\frac{3}{2}=-\frac{27}{20}\)
c) \(\frac{3}{7}-\left(\frac{7}{4}-x\right)=\frac{5}{14}\Leftrightarrow\frac{7}{4}-x=\frac{3}{7}-\frac{5}{14}=\frac{1}{14}\Leftrightarrow x=\frac{7}{4}-\frac{1}{14}=\frac{47}{28}\)
d) \(\left[\frac{8}{3}-\left(-x\right)\right]-\left(\frac{-1}{2}\right)=\frac{5}{3}+\frac{1}{6}=\frac{11}{6}\Rightarrow\frac{8}{3}+x+\frac{1}{2}=\frac{11}{6}\)
\(\Rightarrow\frac{8}{3}+x=\frac{11}{6}-\frac{1}{2}=\frac{4}{3}\Rightarrow x=\frac{4}{3}-\frac{8}{3}=-\frac{4}{3}\)
e) \(\frac{3-x}{9}=\frac{4}{3-x}\Leftrightarrow\left(3-x\right)^2=4.9=36\Rightarrow3-x=6\Leftrightarrow x=3-6=-3\)
Tìm x :
a) \(2.x.\frac{-3}{4}=-\frac{5}{12}\)
\(\Rightarrow2x=-\frac{5}{12}:-\frac{3}{4}\)
\(\Rightarrow2x=\frac{5}{9}\)
\(\Rightarrow x=\frac{5}{9}:2\)
\(\Rightarrow x=\frac{5}{18}\)
Vậy : \(x=\frac{5}{18}\)
b) \(\frac{2}{3}+\frac{1}{3}.x=7\)
\(\Rightarrow\frac{1}{3}.x=7-\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}.x=\frac{19}{3}\)
\(\Rightarrow x=\frac{19}{3}:\frac{1}{3}\)
\(\Rightarrow x=19\)
Vậy : \(x=19\)
c) \(\left(4.x+\frac{1}{8}\right)=\frac{3}{10}\)
\(\Rightarrow4.x=\frac{3}{10}-\frac{1}{8}\)
\(\Rightarrow4.x=\frac{7}{40}\)
\(\Rightarrow x=\frac{7}{40}:4\)
\(\Rightarrow x=\frac{7}{160}\)
Vậy : \(x=\frac{7}{160}\)
d) \(\frac{1}{3}.x-5=1\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}.x-5=\frac{3}{2}\)
\(\Rightarrow\frac{1}{3}.x=\frac{3}{2}+5\)
\(\Rightarrow\frac{1}{3}.x=\frac{13}{2}\)
\(\Rightarrow x=\frac{13}{2}:\frac{1}{3}\)
\(\Rightarrow x=\frac{39}{2}\)
Vậy : \(x=\frac{39}{2}\)
e) \(-\frac{2}{3}.x+\frac{1}{3}=-\frac{1}{2}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{1}{2}-\frac{1}{3}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{5}{6}\)
\(\Rightarrow x=-\frac{5}{6}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=\frac{5}{4}\)
Vậy : \(x=\frac{5}{4}\)
a) = 3/3 x ( -24/54 +45/54 ) : 7/12
= 1 x 21/54 x 12/7
= 18/27
( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )
S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)
S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)
S = \(\frac12\) - \(\frac19\)
S = \(\frac{9}{18}-\frac{2}{18}\)
S = \(\frac{7}{18}\)
a.
| x | + \(\left|-\frac{2}{5}\right|=-\frac{5}{3}\)
| x | + \(\frac{2}{5}=-\frac{5}{3}\)
| x | = \(-\frac{5}{3}-\frac{2}{5}\)
| x | = \(-\frac{31}{15}\)
\(\Rightarrow x\in\varnothing\)vì trị đối \(\ge\)0
Vậy x \(\in\varnothing\)
b.
| x - 3 | = \(\frac{4}{5}\)
\(\Rightarrow\)x - 3 = \(\frac{4}{5}\)hoặc \(-\frac{4}{5}\)
\(\Rightarrow\)x = \(\frac{4}{5}+3\)hoặc \(-\frac{4}{5}+3\)
\(\Rightarrow\)x = \(\frac{19}{5}\)hoặc \(\frac{11}{5}\)
Vậy x \(\in\){ \(\frac{19}{5}\); \(\frac{11}{5}\)}
c.
| x - 7 | = \(\frac{5}{3}\)
\(\Rightarrow\)x - 7 = \(\frac{5}{3}\)hoặc \(-\frac{5}{3}\)
\(\Rightarrow\)x = \(\frac{5}{3}+7\)hoặc \(-\frac{5}{3}+7\)
\(\Rightarrow\)x = \(\frac{26}{3}\)hoặc \(\frac{16}{3}\)
Vậy x \(\in\){ \(\frac{26}{3}\); \(\frac{16}{3}\)}
d.
| x - \(\frac{1}{2}\)| = \(\frac{1}{4}\)
\(\Rightarrow\)x - \(\frac{1}{2}\)= \(\frac{1}{4}\)hoặc \(-\frac{1}{4}\)
\(\Rightarrow\)x = \(\frac{1}{4}+\frac{1}{2}\)hoặc \(-\frac{1}{4}+\frac{1}{2}\)
\(\Rightarrow\)x = \(\frac{3}{4}\)hoặc \(\frac{1}{4}\)
Vậy x \(\in\){ \(\frac{3}{4}\); \(\frac{1}{4}\)}
e.
| x - 7 | = \(-\frac{5}{3}\)
\(\Rightarrow\)x - 7 = \(-\frac{5}{3}\)hoặc \(\frac{5}{3}\)
\(\Rightarrow\)x = \(-\frac{5}{3}+7\)hoặc \(\frac{5}{3}+7\)
\(\Rightarrow\)x = \(\frac{16}{3}\)hoặc \(\frac{26}{3}\)
Vậy x \(\in\){ \(\frac{16}{3}\); \(\frac{26}{3}\)}
\(a) |x|+ |-\frac{2}{5}|=|-\frac{5}{3}|\)
\(|x|+\frac{2}{5}=\frac{5}{3}\)
\(|x|=\frac{5}{3}-\frac{2}{5}\)
\(|x|=\frac{19}{15}\)
\(=> x = \frac{19}{15}\) hoặc \(-\frac{19}{15}\)
\(b) |x-3|=\frac{4}{5}\)
Ta có 2 trường hợp :
\(TH1 : x-3=\frac{4}{5}\)
\(x=\frac{4}{5}+3\)
\(x=\frac{19}{5}\)
\(TH2 : x- 3 = -\frac{4}{5}\)
\(x=-\frac{4}{5}+3\)
\(x=\frac{11}{5}\)
\(c) |x-7|=\frac{5}{3}\)
\(TH1:x-7=\frac{5}{3} \)
\(x =\frac{5}{3}+7 \)
\( x=\frac{26}{3}\)
\(TH2 : x-7=-\frac{5}{3}\)
\(x=-\frac{5}{3}+7 \)
\(x=\frac{16}{3}\)
\(d) |x-\frac{1}{2}|=\frac{1}{4}\)
\(TH1 : x-\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}+\frac{1}{2}\)
\(x=\frac{3}{4}\)
\(TH2 : x-\frac{1}{2}=-\frac{1}{4}\)
\(x=-\frac{1}{4}+\frac{1}{2} \)
\( x=\frac{1}{4}\)
\(e)|x-7|=-\frac{5}{3}\)
\(TH1:x-7=\frac{5}{3}\)
\(x =\frac{5}{3}+7\)
\(x=\frac{26}{3}\)
\(TH2 : x-7=-\frac{5}{3}\)
\(x=-\frac{5}{3}+7\)
\(x=\frac{16}{3}\)