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a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
Giải:
a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{11}\)
\(2x=\dfrac{16}{11}-\dfrac{7}{2}\)
\(2x=\dfrac{-45}{22}\)
\(x=\dfrac{-45}{22}:2\)
\(x=\dfrac{-45}{44}\)
b) \(3-\left(17-x\right)=-12\)
\(3-17+x=-12\)
\(x=-12-3+17\)
\(x=2\)
c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}:\dfrac{2}{3}\)
\(x=\dfrac{-3}{5}\)
d) \(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}\)
\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}:2\)
\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)
Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\)
e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\)
\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\)
\(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\)
\(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\)
\(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\)
\(-0,6x=\dfrac{-7}{18}\)
\(x=\dfrac{-7}{18}:-0.6\)
\(x=\dfrac{35}{54}\)
f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\)
\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\)
\(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\)
\(\dfrac{3}{5}.x=\dfrac{13}{9}\)
\(x=\dfrac{13}{9}:\dfrac{3}{5}\)
\(x=\dfrac{65}{27}\)
Chúc bạn học tốt!
\(\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{1}{3}\\2x-\dfrac{2}{3}=\dfrac{-1}{3}\end{matrix}\right.\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
a)\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)
=>2x=-3/2
hay x=-3/4
b: 2x+3=5
=>2x=2
hay x=1
c: =>3(x-2)=4(5+x)
=>4x+20=3x-6
=>x=-26
a) 2x - 17 = 35 : 32
2x - 17 = 35 - 2
2x - 17 = 33
= 27
2x = 27 +17
2x = 44
x = 44 :2
vậy x = 22
b, ( 19 - x ) .2 - 20 = 23
= 8
( 19 - x ) .2 = 8 + 20
( 19 - x ) .2 = 28
19 - x = 28 :2
19 - x = 14
x = 19 - 14
vậy x = 5
a)2x-17=243:9
2x-17=27
2x=27+17
2x=44
vậy x=44
b)(19-x).2-20=8
(19-x).2=20+8=28
19-x=28:2
19-x=14
x=19-14
x=5
vậy x=5
c)Bạn tách 3^2 thành 3^1.3^2 nhá . Tương tự tách các phần rồi rút gọn là xong .mik nhắn trên máy tính nó khó ko nhanh được ý ạ bạn thông ca,r
CHUC BAN HOC TOT