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`x :3*5 = 3/4 :(-5/6)`
`x :15 =3/4*(-6/5)=-9/10`
`x = -9/10 *15 =-27/2`
`x-1*2/2 = 8/x -1.2`
`x- 1*1 = 8/x -2`
`x-8/x = -2+1`
`x-8/x =-1`
`x^2 -8x =-x`
`x^2 -8x +x=0`
`x^2 -7x =0`
`x(x-7) =0`
`=>[(x=0),(x=7):}`
`a, x \div 15=-9/10`
`x=-9/10*14`
`x=-27/2`
`b, (x-1*2)/2=8/(x-1*2)`
\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)
`(x-1*2)^2=16`
`(x-1*2)^2=(+-4)^2`
\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
2(x4+3)-(9)=17
⇒2x4+6+9=17
⇒2x4+15=17
⇒ 2x4=2
⇒ x4=1
⇒ x=\(\pm1\)
b) 5x2.x+1-3.42=-47
⇒5x3+1-48=-47
⇒5x3-47=-47
⇒5x3=0
⇒x3=0
⇒x=0
a) \(2\left(x^4+3\right)-\left(-9\right)=17\)
\(2x^4+6+9=17\)
\(2x^4=2\)
\(x^4=1\)
⇒ \(x=1\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)