Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
a) x=7-\dfrac{2}{5}+1,62=8,22x=7−52+1,62=8,22
b) x=4 \dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4 \dfrac{3}{10}x=453+51−21=4103
c) 2 x-x=\dfrac{3}{5}+\dfrac{4}{7}2x−x=53+74
x=\dfrac{41}{35}x=3541
d) x=3 \dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0.25x=321−75+131−0.25
x=2 \dfrac{223}{364}x=2364223
x=7-\dfrac{2}{5}+1,62=8,22x=7−52+1,62=8,22
b) x=4 \dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4 \dfrac{3}{10}x=453+51−21=4103
c) 2 x-x=\dfrac{3}{5}+\dfrac{4}{7}2x−x=53+74
x=\dfrac{41}{35}x=3541
d) x=3 \dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0.25x=321−75+131−0.25
x=2 \dfrac{223}{364}x=2364223
a, \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{4}{5}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}\\ \Rightarrow x=\dfrac{3}{10}\cdot\dfrac{3}{2}\\ \Rightarrow x=\dfrac{9}{20}\)
b, \(x+\dfrac{1}{4}=\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{4}{3}-\dfrac{1}{4}\\ \Rightarrow x=\dfrac{13}{12}\)
c, \(\dfrac{3}{5}x-\dfrac{1}{2}=-\dfrac{1}{7}\)
\(\Rightarrow\dfrac{3}{5}x=-\dfrac{1}{7}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{5}x=-\dfrac{9}{14}\\ \Rightarrow x=-\dfrac{9}{14}\cdot\dfrac{5}{3}\\ \Rightarrow x=\dfrac{15}{14}\)
d, \(\left|x-\dfrac{4}{5}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=-\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}+\dfrac{4}{5}\\x=-\dfrac{3}{4}+\dfrac{4}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)
e, \(lxl\) là j mk ko hiểu!
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
a) x + 2/5 = -4/3
x = -4/3 - 2/5
x = -26/15
b) -5/6 + 1/3 x = (-1/2)²
-5/6 + 1/3 x = 1/4
1/3 x = 1/4 + 5/6
1/3 x = 13/12
x = 13/12 : 1/3
x = 13/4
c) 7/12 - (x + 7/6) . 6/5 = (-1/2)³
7/12 - (x + 7/6) . 6/5 = -1/8
(x + 7/6) . 6/5 = 7/12 + 1/8
(x + 7/6) . 6/5 = 17/24
x + 7/6 = 17/24 : 6/5
x + 7/6 = 85/144
x = 85/144 - 7/6
x = -83/144
\(a,x+\dfrac{2}{5}=-\dfrac{4}{3}\\ \Rightarrow x=-\dfrac{26}{15}\\ b,-\dfrac{5}{6}+\dfrac{1}{3}x=\left(-\dfrac{1}{2}\right)^2\\ \Rightarrow-\dfrac{5}{6}+\dfrac{1}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{3}x=\dfrac{13}{12}\\ \Rightarrow x=\dfrac{13}{4}\\ c,\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\left(-\dfrac{1}{2}\right)^3\\ \Rightarrow\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=-\dfrac{1}{8}\\ \Rightarrow\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\dfrac{17}{24}\\ \Rightarrow x+\dfrac{7}{6}=\dfrac{85}{144}\\ \Rightarrow x=-\dfrac{83}{144}.\)
a, \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy x = 23
b, \(\dfrac{x+4}{20}=\dfrac{5}{5x+4}\)
\(\Rightarrow\left(x+4\right)\left(5x+4\right)=100\)
\(\Rightarrow5x^2+24x+16=100\)
sai đề à?
c, \(\dfrac{7}{x+1}=\dfrac{x+1}{9}\)
\(\Rightarrow\left(x+1\right)^2=63\)
\(\Rightarrow\left[{}\begin{matrix}x+1=\sqrt{63}\\x+1=-\sqrt{63}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{63}-1\\x=-\sqrt{63}-1\end{matrix}\right.\)
Vậy...
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
a, \(\dfrac{5}{2}\)\(x\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{5}{2}\)\(x\) = \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
\(\dfrac{5}{2}\)\(x\) = 1
\(x\) = 1: \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{2}{5}\)
b, \(\dfrac{x+4}{20}\) = \(\dfrac{5}{x+4}\) (đk \(x\) ≠ -4)
(\(x\)+4).(\(x\) + 4) = 20.5
(\(x\)+ 4)2 = 100
(\(x\) + 4)2 = 102
\(\left[{}\begin{matrix}x+4=-10\\x+4=10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-10-4\\x=10-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-14\\x=6\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-14; 6}