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mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
a. \(\dfrac{3}{4}-\left(2x-\dfrac{2}{3}\right)=\dfrac{-5}{6}\)
\(\Rightarrow2x-\dfrac{2}{3}=\dfrac{3}{4}-\dfrac{-5}{6}\)
\(\Rightarrow2x-\dfrac{2}{3}=\dfrac{19}{12}\)
\(\Rightarrow2x=\dfrac{19}{12}+\dfrac{2}{3}=\dfrac{9}{4}\)
\(\Rightarrow x=\dfrac{9}{4}:2=\dfrac{9}{8}\)
Vậy............
b. \(1,5-\left(x+\dfrac{7}{2}\right)=2^7:2^5\)
\(\Rightarrow1,5-\left(x+\dfrac{7}{2}\right)=2^2=4\)
\(\Rightarrow x+\dfrac{7}{2}=1,5-4=\dfrac{-5}{2}\)
\(\Rightarrow x=\dfrac{-5}{2}-\dfrac{7}{2}=-6\)
Vậy.............
a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)
\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)
=>2x+5=1/44 hoặc 2x+1=-1/44
=>x=-219/88 hoặc x=-221/88
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
Bài 1:
a) \(x^2-3=1\)
\(\Rightarrow x^2=1+3=4\)
\(\Rightarrow x=\pm2\)
b)\(2x^3+12=-4\)
\(\Rightarrow2x^3=-4-12=-16\)
\(\Rightarrow x^3=-8\)
\(\Rightarrow x=-2\)
c)\(\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)
b) \(2x^3+12=-4\Rightarrow2x^3=-16\)
\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)
\(\Rightarrow x=-2\)
c) \(\left(2x-3\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d,h,i,k cững tương tự....
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a,\(x:\left(\dfrac{-3}{5}\right)^2=\dfrac{-3}{5}\)
\(\Leftrightarrow x=\dfrac{-3}{5}.\dfrac{9}{25}\Leftrightarrow x=\dfrac{-27}{125}\)
b,\(\left(\dfrac{-1}{3}\right)^3.x=\dfrac{1}{81}\)
\(\Leftrightarrow x=.\dfrac{1}{81}:\left(\dfrac{-1}{27}\right)\Leftrightarrow x=\dfrac{-1}{3}\)
c,(2x2)=16\(\Leftrightarrow\)x2=8\(\Leftrightarrow\)x=\(,\sqrt{8}\)
Giải:
a. \(x:\left(\dfrac{-3}{5}\right)^2=\dfrac{-3}{5}\)
\(\Rightarrow x=\left(\dfrac{-3}{5}\right)^2.\dfrac{-3}{5}\)
\(\Rightarrow x=\left(\dfrac{-3}{5}\right)^3\)
\(\Rightarrow x=\dfrac{-27}{125}\)
Vậy.................
b.\(\left(\dfrac{-1}{3}\right)^3.x=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)^3\)
\(\Rightarrow x=\dfrac{1}{81}:\dfrac{-1}{27}\)
\(\Rightarrow x=\dfrac{1}{81}.\dfrac{-27}{1}\)
\(\Rightarrow x=\dfrac{1}{3}.\dfrac{-1}{1}\)
\(\Rightarrow x=\dfrac{-1}{3}\)
vậy................
c.\(\left(2x^2\right)=16\)
\(\Rightarrow2x=\sqrt{16}=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
vậy....................
*** \(\dfrac{15-x}{8}=\dfrac{x-23}{10}\)
\(\Rightarrow10\left(15-x\right)=8\left(x-23\right)\)
\(\Rightarrow150-10x=8x-184\)
\(\Rightarrow150+184=10x+8x\)
\(\Rightarrow18x=334\)
\(\Rightarrow x=\dfrac{167}{9}\)
*** \(\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right).\left(2015\right)^0\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right).1\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|=\left(-\dfrac{1}{2}\right)+3\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|=\dfrac{29}{10}\)
\(\Rightarrow\left|2x-1\right|=\dfrac{29}{10}:\dfrac{1}{2}\)
\(\Rightarrow\left|2x-1\right|=\dfrac{29}{5}\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{29}{5}\\2x-1=-\dfrac{29}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{34}{5}\\2x=-\dfrac{24}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Bài 2 từ dòng 2 đến dòng 7 nên dùng dấu \(\Leftrightarrow \) mới đúng em nhé.
1a, (-3,8) + [ (-5,7)+3,8]
= (-3,8+3,8) -5,7
= -5,7
b, 31,4 +[6,4 +(-18)]
= 31,4 - 11,6
= 19,8
c,[(9,6 )+4,5 ] +[9,6 +(-1,5)]
= 9,6 +9,6 +(4,5-1,5)
= 19,2 +3
=22,2
d, [ (-4,9) + (-7,8)] +[1,9+2,8]
= (-4,9 +1,9)+ (-7,8+2,8)
= -3 -5
=-8
e, (3,1-2,5)-(-2,5+3,1)
= (3,1-3,1)+(2,5-2,5)
=0
f, (5,3 -2,8 )- (4+5,3)
= (5,3-5,3) -2,8-4
= -6,8
g, -(251,3 +281 )+ 3,251 -(1-281)
= -251,3-281+3,251 -1+281
= (-281+281) -251,3 +3,251 -1
= -249,049
h,-(3/54+3/4)-(-3/4 +2/5)
= -3/54-3/4 +3/4 +2/5
=(-3/4+3/4) -3/54 +2/5
=123/270
chúc bạn học tốt
|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 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)