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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....
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:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
|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) \(x:\left(-\dfrac{1}{2}\right)^3=-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}.\left(-\dfrac{1}{2}\right)^3\)
\(x=-\dfrac{1}{2}^{1+3}\)
\(x=-\dfrac{1}{2}^4\)
\(x=\dfrac{\left(-1\right)^4}{2^4}\)
\(x=\dfrac{1}{16}\)
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) \(\dfrac{2}{5}\)-\(\left(\dfrac{1}{10}-x\right)\)=\(\left(\dfrac{-2}{5}-\dfrac{1}{2}\right)^2\)
\(\dfrac{2}{5}\)- \(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{2}{5}\)-\(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)=\(\dfrac{7}{20}\)
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) \(2.\left(x-\dfrac{1}{3}\right)-3\left(x-\dfrac{1}{2}\right)=\dfrac{1}{2}x\)
\(\Leftrightarrow2x-\dfrac{2}{3}-3x+\dfrac{3}{2}=\dfrac{1}{2}x\)
\(\Leftrightarrow-\dfrac{2}{3}+\dfrac{3}{2}+2x-3x=\dfrac{1}{2}x\)
\(\Leftrightarrow\dfrac{5}{6}-1x=\dfrac{1}{2}x\)
\(\Leftrightarrow-\dfrac{3}{2}x=\dfrac{5}{6}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
Vậy x=-5/9
b) Cách làm tương tự nhé! Bạn nhập bài toán trên vào rồi bấm SHIFT CALC= sẽ nhanh hơn :))
Chúc bạn học tốt!
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) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)
Lập bảng xét dấu:
| x | -2 | \(\dfrac{1}{2}\) | |||
| x + 2 | - | 0 | + | + | |
| x - \(\dfrac{1}{2}\) | - | - | 0 | + |
TH : Xét x < -2
Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)
-2x = 2\(\dfrac{1}{4}\)
=> x = \(-1\dfrac{1}{8}\) ( loại )
TH 2: \(-2\le x< \dfrac{1}{2}\)
Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
=> \(2,5=\dfrac{3}{4}\) ( loại )
TH3 : \(x\ge\dfrac{1}{2}\)
x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
2x + 1,5 = \(\dfrac{3}{4}\)
x = -0,375( loại )
vậy ....
b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)
c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)
TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)
\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)
TH2 : \(x-1< 0\Rightarrow x< 1\)
=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)
Vậy x = 1
\(2\left(x+1\dfrac{1}{3}\right)=\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{6}\\ x+\dfrac{4}{3}=\dfrac{1}{6}:2\\ x+\dfrac{4}{3}=\dfrac{1}{12}\\ x=\dfrac{1}{12}-\dfrac{4}{3}\\ x=-\dfrac{15}{12}=\dfrac{-5}{4}\)
\(2\left(x+1\dfrac{1}{3}\right)=\left(-\dfrac{1}{2}\right)^2\cdot\dfrac{2}{3}\)
=>\(2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}=\dfrac{1}{6}\)
=>\(x+\dfrac{4}{3}=\dfrac{1}{12}\)
=>\(x=\dfrac{1}{12}-\dfrac{4}{3}=\dfrac{1}{12}-\dfrac{16}{12}=-\dfrac{15}{12}=-\dfrac{5}{4}\)