
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
Vậy x = -10
b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Rightarrow4,5-2x=\dfrac{121}{98}\)
\(\Rightarrow2x=\dfrac{160}{49}\)
\(\Rightarrow x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(2,8x-32=-90.\dfrac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy \(x=-10\)
\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)
\(4,5-2x=\dfrac{121}{98}\)
\(2x=4,5-\dfrac{121}{98}\)
\(2x=\dfrac{160}{49}\)
\(x=\dfrac{160}{49}:2\)
\(x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)

\(4,5-2x.1\dfrac{4}{7}=\dfrac{11}{14}\)
<=>\(-\dfrac{22}{7}x=\)\(\dfrac{11}{14}-4,5\)
<=>\(-\dfrac{22}{7}x=-\dfrac{26}{7}\)
<=>\(x=-\dfrac{26}{7}:\left(-\dfrac{22}{7}\right)\)
<=>\(x=\dfrac{13}{11}\)
4,5 -2x.11/7=11/4
2x.11/7=4.5-11/4
2x.11/7=7/4
2x =7/4: 11/7
2x=49/44
x=49/44: 2
x=49/88

a, 4,5 - 2x . 147 = 1114
2x . 147 = 4,5 - 1114
2x . 147 = -1109,5
2x = -1109,5 / 147
2x = \(-\dfrac{317}{42}\)
x = \(-\dfrac{317}{42}:2\)
x = \(-\dfrac{317}{84}\)
b, ( x + 14 - 13 ) : ( 2 + 16 - 14 ) = 746
( x + 1 ) : 4 = 746
x + 1 = 746 . 4
x + 1 = 2984
x = 2984 - 1
x = 2983
c, \(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{1}{6}\)
\(\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\)
\(\dfrac{13}{21}+x=\dfrac{2}{7}\)
\(x=\dfrac{2}{7}-\dfrac{13}{21}\)
\(x=-\dfrac{1}{3}\)

a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{32}=\dfrac{2}{x+1}\left(đk:x\ne1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=64\)
\(\Leftrightarrow\left(x+1\right)^2-64=0\)
\(\Leftrightarrow x^2+2x+1-64=0\)
\(\Leftrightarrow x^2+6x-63=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+16}{2}\\x=\dfrac{-2-16}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\left(đk:x\ne-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)
Vậy \(x_1=-9;x_2=7\)
b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\Leftrightarrow\dfrac{x+1}{5}=\dfrac{7}{x-1}\left(đk:x\ne1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=35\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)-35=0\)
\(\Leftrightarrow x^2-1-35=0\)
\(\Leftrightarrow x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\left(đk:x\ne1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Vậy \(x_1=-6;x_2=6\)
c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)
\(\Leftrightarrow\left|4,5-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|4,5-2x\right|\cdot\dfrac{4}{11}=\dfrac{11}{14}\)
\(\Leftrightarrow\dfrac{4}{11}\cdot\left|4,5-2x\right|=\dfrac{11}{14}\)
\(\Leftrightarrow\left|4,5-2x\right|=\dfrac{121}{56}\)
\(\Leftrightarrow\left[{}\begin{matrix}4,5-2x=\dfrac{121}{56}\\4,5-2x=-\dfrac{121}{56}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{131}{112};x_2=\dfrac{373}{112}\)
a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)=32.2\)
\(\Rightarrow\left(x+1\right)^2=64\)
\(\Rightarrow\left(x+1\right)^2=8^2\)
\(\Rightarrow x+1=8\)
\(\Rightarrow x=8-1\)
\(\Rightarrow x=7\left(TM\right)\)
Vậy \(x=7\) là giá trị cần tìm
b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\Rightarrow\left(x+1\right)\left(x-1\right)=7.5\)
\(\Rightarrow\left[{}\begin{matrix}x+1=7\\x-1=5\end{matrix}\right.\) \(\Rightarrow x=6\left(TM\right)\)
Vậy \(x=6\) là giá trị cần tìm
c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)
\(\left|\dfrac{45}{10}-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)
\(\left|\dfrac{9}{2}-2x\right|=\dfrac{11}{14}.\dfrac{11}{4}\)
\(\left|\dfrac{9}{2}-2x\right|=\dfrac{121}{56}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{9}{2}-2x=\dfrac{121}{56}\\\dfrac{9}{2}-2x=\dfrac{-121}{56}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{131}{56}\\2x=\dfrac{373}{56}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{131}{112};\dfrac{373}{112}\right\}\) là giá trị cần tìm

A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)

Anh làm lại câu b)
\(\left(4,5-200\%x\right).1\dfrac{4}{7}=\dfrac{11}{14}\\ < =>\left(\dfrac{9}{2}-2x\right).\dfrac{11}{7}=\dfrac{11}{14}\\ =>\dfrac{9}{2}-2x=\dfrac{\dfrac{11}{14}}{\dfrac{11}{7}}=\dfrac{1}{2}\\ =>2x=\dfrac{9}{2}-\dfrac{1}{2}=4\\ =>x=\dfrac{4}{2}=2\)
a, \(3x+\dfrac{1}{8}=2\dfrac{3}{4}\\ < =>3x+\dfrac{1}{8}=\dfrac{11}{4}\\ =>3x=\dfrac{11}{4}-\dfrac{1}{8}=\dfrac{21}{8}\\ =>x=\dfrac{\dfrac{21}{8}}{3}=\dfrac{7}{8}\)
b, \(\left(4,5-200\%x\right).1\dfrac{4}{7}=\dfrac{11}{14}\\ < =>\left(4,5-2x\right).\dfrac{11}{7}=\dfrac{11}{4}\\ =>4,5-2x=\dfrac{11}{4}:\dfrac{11}{7}=\dfrac{7}{4}\\ =>2x=4,5-\dfrac{7}{4}=\dfrac{11}{4}\\ =>x=\dfrac{\dfrac{11}{4}}{2}=\dfrac{11}{8}\)

a) \(\left|x+3\right|:\left(-15\right)=\frac{1}{3}\)
\(\left|x+3\right|=-5\)
=> không tìm được x
b) \(\left|4,5-2x\right|.\left(-1\frac{4}{7}\right)=-\frac{11}{14}\)
\(\left|4,5-2x\right|=2\)
TH1: 4,5 - 2x = 2
2x = 2,5
x = 1,25
TH2: 4,5 - 2x = -2
2x = 6,5
x = 3,25
KL:...

\(\left(4,5-2x\right).\frac{14}{7}=\frac{11}{4}\)
\(4,5-2x=\frac{11}{4}:\frac{14}{7}\)
\(4,5-2x=\frac{77}{56}\)
\(2x=\frac{77}{56}+4,5\)
\(2x=\frac{77}{56}+\frac{9}{2}\)
\(2x=\frac{329}{56}\)

1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
Ta có : 4,5 - 2x * 147 = 11/14 .
⇒ 9/2 - 2x * 147 = 11/14 .
⇒ 2x * 147 = 9/2 - 11/14 .
⇒ 2x * 147 = 63/14 - 11/14 .
⇒ 2x * 147 = 26/7 .
⇒ 2x = 26/7 : 147 .
⇒ 2x = ( 26 * 147 ) / 7 .
⇒ 2x = 78 .
⇒ x = 78 : 2 .
⇒ x = 39 .
Vậy x = 39 .
\(4,5-2x\cdot147=\dfrac{11}{14}\\ 2x\cdot147=4,5-\dfrac{11}{14}\\ 2x\cdot147=\dfrac{52}{14}\\ 2x=\dfrac{52}{14}:147\\ 2x=\dfrac{26}{1029}\\ x=\dfrac{26}{1029}:2\\ x=\dfrac{13}{1029}\\ Vậy...\)