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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 :((
Câu a:
\(\frac{x+5}{x+9}\) = \(\frac{x+4}{x+3}\)
(\(x\) + 5)(\(x+3\)) = (\(x+4\))(\(x+9\))
\(x^2\) + 3\(x\) + 5\(x\) + 15 = \(x^2\) + 9\(x\) + 4\(x\) + 36
\(x^2+3x+5x\) - \(x^2-9x\) - 4\(x\) = 36 - 15
(\(x^2-x^2\)) + (\(3x+5x\) - 9\(x-4x\)) = 21
0 + (8\(x\) - 9\(x\) - 4\(x\)) = 21
-\(x-4x\) = 21
-5\(x\) = 21
\(x\) = 21 : (-5)
\(x\) = - \(\frac{21}{5}\)
Vậy \(x=-\frac{21}{5}\)
0,7.2\(\frac23\).20.0,375.\(\frac{5}{28}\)
= \(\frac{7}{10}\) .\(\frac83\).20.\(\frac38\).\(\frac{5}{28}\)
= (\(\frac{7}{10}\).\(\frac{5}{28}\).20).(\(\frac83\).\(\frac38\))
= (\(\frac18.20\)). 1
= \(\frac52\)
B = (9,75 .21\(\frac37\) + \(\frac{39}{4}\).18\(\frac47\))
B = (\(\frac{39}{4}\).\(\frac{150}{7}\frac{}{7}\)+ \(\frac{39}{4}\).\(\frac{130}{7}\))
B = \(\frac{39}{4}\).(\(\frac{150}{7}\) + \(\frac{130}{7}\))
B = \(\frac{39}{4}\).\(\frac{280}{7}\)
B = 390
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
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x\) = \(\frac{1}{10}+\frac{1}{2}\)
\(\frac{2}{3}x\) = \(\frac{3}{5}\)
\(x\) = \(\frac{3}{5}:\frac{2}{3}\)
\(x\) = \(\frac{9}{10}\)
Câu b:
(2 4/5x - 50) : 2/3 = 51
2 4/5x - 50 = 51 x 2/3
14/5x - 50 = 34
14/5x = 34 + 50
14/5x = 84
x = 84 : 14/5
x = 30
Vậy x = 30
|\(\frac32x\) + \(\frac12\)| = |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 \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
Bài 2:
b: x+25%x=-1,25
=>1,25x=-1,25
hay x=-1
c: x-75%x=1/4
=>1/4x=1/4
hay x=1
Bài 2:
a: =3/2-11/4=6/4-11/4=-5/4
b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3

a) \(\frac{a}{3b}\)= \(\frac{c}{3d}\)
Đặt \(\frac{a}{b}\)= \(\frac{c}{d}\)= k
=.> a = b.k, c = d.k
Từ đó ta có
+) \(\frac{a}{c-3b}\)= \(\frac{bk}{3k-3b}\)= \(\frac{bk}{b.\left(k-3\right)}\)= \(\frac{k}{k-3}\) (1)
+) \(\frac{c}{c-3d}\)= \(\frac{dk}{dk-3d}\)-= \(\frac{dk}{d.\left(k-3\right)}\)= \(\frac{k}{k-3}\) (2)
Từ (1) và (2) ta có
\(\frac{a}{a-3b}\) = \(\frac{c}{c-3d}\)