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|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 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
*** \(\dfrac{15-x}{8}=\dfrac{x-23}{10}\)
\(\Rightarrow10\left(15-x\right)=8\left(x-23\right)\)
\(\Rightarrow150-10x=8x-184\)
\(\Rightarrow150+184=10x+8x\)
\(\Rightarrow18x=334\)
\(\Rightarrow x=\dfrac{167}{9}\)
*** \(\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right).\left(2015\right)^0\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right).1\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|=\left(-\dfrac{1}{2}\right)+3\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|=\dfrac{29}{10}\)
\(\Rightarrow\left|2x-1\right|=\dfrac{29}{10}:\dfrac{1}{2}\)
\(\Rightarrow\left|2x-1\right|=\dfrac{29}{5}\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{29}{5}\\2x-1=-\dfrac{29}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{34}{5}\\2x=-\dfrac{24}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Bài 2 từ dòng 2 đến dòng 7 nên dùng dấu \(\Leftrightarrow \) mới đúng em nhé.
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) \(x:\left(-\dfrac{1}{2}\right)^3=-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}.\left(-\dfrac{1}{2}\right)^3\)
\(x=-\dfrac{1}{2}^{1+3}\)
\(x=-\dfrac{1}{2}^4\)
\(x=\dfrac{\left(-1\right)^4}{2^4}\)
\(x=\dfrac{1}{16}\)
b) \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(x=\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^5\)
\(x=\left(\dfrac{3}{4}\right)^{7-5}\)
\(x=\left(\dfrac{3}{4}\right)^2\)
\(x=\dfrac{3^2}{4^2}\)
\(x=\dfrac{9}{16}\)
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....................
a) \(\dfrac{2}{5}\)-\(\left(\dfrac{1}{10}-x\right)\)=\(\left(\dfrac{-2}{5}-\dfrac{1}{2}\right)^2\)
\(\dfrac{2}{5}\)- \(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{2}{5}\)-\(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)=\(\dfrac{7}{20}\)
x = \(\dfrac{1}{10}\)-\(\dfrac{7}{20}\)
x = \(\dfrac{-1}{4}\)
Chúc bn học tốt
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,
\(\dfrac{1}{4}x-1+\dfrac{1}{3}\left(\dfrac{5}{2}x-7\right)-\left(\dfrac{5}{8}x-2\right)=\dfrac{7}{2}\)
\(\Rightarrow\dfrac{1}{4}x-1+\dfrac{5}{6}x-\dfrac{7}{3}-\dfrac{5}{8}x+2=\dfrac{7}{2}\)
\(\Rightarrow\dfrac{1}{4}x+\dfrac{5}{6}x-\dfrac{5}{8}x=\dfrac{7}{2}+1+\dfrac{7}{3}-2\)
\(\Rightarrow\dfrac{11}{24}x=\dfrac{29}{6}\)
\(\Rightarrow x=\dfrac{116}{11}\)
b,
\(\left|2-\dfrac{3}{2}x\right|-4=x+2\)
\(\Rightarrow\left|2-\dfrac{3}{2}x\right|=x-2\)
\(\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-\left(x+2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-2=x+\dfrac{3}{2}x\\2+2=-x+\dfrac{3}{2}x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=0\\\dfrac{1}{2}x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
c,
\(-3\left(\dfrac{2}{5}x-\dfrac{1}{5}\right)-x\left(x-\dfrac{1}{2}\right)=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{6}{5}x+\dfrac{3}{5}-x^2+\dfrac{1}{2}x=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{7}{10}x=\dfrac{1}{6}-\dfrac{3}{5}-x^2+x^2\)
\(\Rightarrow-\dfrac{7}{10}x=-\dfrac{13}{30}\Leftrightarrow x=\dfrac{13}{21}\)
Ta có: \(3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{15}\)
\(\Leftrightarrow3x-5x+x=\dfrac{1}{15}+\dfrac{3}{2}+3\)
\(\Leftrightarrow x=-\dfrac{137}{30}\)