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a: =>5/42-x=11/13-15/28+11/13=421/364
=>x=-1193/1092
b: =>\(\dfrac{7}{2}-2x=7+\dfrac{6}{5}-3-\dfrac{2}{5}-1-\dfrac{4}{5}=3\)
=>2x=1/2
=>x=1/4
c: =>|2x-1/3|=-1/3(vô lý)
d: =>2x-1=-3
=>2x=-2
hay x=-1
e: =>2x=16
hay x=8
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)
\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)
Toàn câu dễ nên bạn tự làm đi.
Trong lúc bạn đánh xong bài này thì bạn có thể làm xong rồi đó.
Đừng có ỷ lại vào người khác ,động não lên.
a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)
\(x=\dfrac{8}{13}-\dfrac{3}{4}\)
\(x=-\dfrac{7}{52}\)
b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
c, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(2x-\dfrac{1}{7}=0\)
\(x-\dfrac{1}{7}=0:2\)
\(x-\dfrac{1}{7}=0\)
\(x=0-\dfrac{1}{7}\)
\(x=\dfrac{1}{7}\)
d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)
\(1:x=\dfrac{2}{5}\)
\(x=1:\dfrac{2}{5}\)
\(x=\dfrac{5}{2}\)
a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)
c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)
\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)
\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)
vậy \(x=0;x=\dfrac{1}{7}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
*Trả lời :
a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
=\(\dfrac{3}{4}.\left(-8\right)\)
= \(-6\)
b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
=\(10:\left(-\dfrac{5}{7}\right)\)
=\(-14\)
c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )
=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)
=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)
=\(\left(-1+1\right):\dfrac{7}{11}\)
\(=0:\dfrac{7}{11}\)
=0.
d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
a. \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Rightarrow\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{-113}{364}\right)=\dfrac{113}{364}\)
\(\Rightarrow\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}-\dfrac{113}{364}\)
\(\Rightarrow\left(\dfrac{5}{42}-x\right)=\dfrac{15}{28}\)
\(\Rightarrow x=\dfrac{5}{42}-\dfrac{15}{28}=\dfrac{-5}{12}\)
Vậy..............
b. \(2x.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{7}\end{matrix}\right.\)
Vậy............
c. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)
Vậy...........
a) C = 20013 - |5−2x|
do \(-\left|5-2x\right|\le0\forall x\)
=> 20013-\(\left|5-2x\right|\le20013\)
=>A≤20013
=> GTLN C =20013 khi 5-2x=0
=> 2x=5
=> x=\(\dfrac{5}{2}\)
vậy GTLN C = 20013 khi x=\(\dfrac{5}{2}\)
b) D = 7 - \(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\)
do \(-\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le0\forall x\)
=> 7-\(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le7\)
=> D≤7
=> GTLN D =7 khi \(\dfrac{2}{3}+\dfrac{1}{4}x=0\)
=> x=-\(\dfrac{8}{3}\)
\(13\dfrac{1}{3}:1\dfrac{1}{3}=26:\left(2x-1\right)\)
\(\Rightarrow\dfrac{40}{3}:\dfrac{4}{3}=26:\left(2x-1\right)\)
\(\Rightarrow\dfrac{40}{3}\cdot\dfrac{3}{4}=26:\left(2x-1\right)\)
\(\Rightarrow\dfrac{40}{4}=26:\left(2x-1\right)\)
\(\Rightarrow26:\left(2x-1\right)=10\)
\(\Rightarrow2x-1=\dfrac{13}{5}\)
\(\Rightarrow2x=\dfrac{13}{5}+1\)
\(\Rightarrow2x=\dfrac{18}{5}\)
\(\Rightarrow x=\dfrac{18}{5}:2\)
\(\Rightarrow x=\dfrac{9}{5}\)
13 13:113=26:(2x−1)\dfrac{1}{3}:1\dfrac{1}{3}=26:\left(2x-1\right)
403:
43 = 26 : ( 2x-1)
403. 34 = 26 : (2x -1)
10= 26: ( 2x -1 )
2x -1 = 26: 10
2x-1= 2,6
2x = 2,6 +1
2x = 3,6
x = 3,6 : 2
x = 1,8.
À mà em mới lớp 6 thôi