Tìm x biết:
a, 8<\(2^9\).\(2^{-5}\) b, 27<\(81^3:3^x< 243\)
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a: \(13329:x=3\)
=>\(x=\dfrac{13329}{3}=4443\)
b: x:8=5678
=>\(x=5678\cdot8=45424\)
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
\(a,\left(8+x\right)\left(6-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\\ b,x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
a) (8+x).(6-x)=0
<=> 8+x = 0 hoặc 6-x = 0
=> x = -8 hoặc x = 6
b) c) x^2 - 5x=0
<=> x^2 = 0 hoặc -5x = 0
=> x = 0 hoặc x = 5
`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`
\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
\(a,\Rightarrow x=19-17=2\\ b,\Rightarrow x+8=28:2=14\\ \Rightarrow x=14-8=6\\ c,\Rightarrow42-x=5^2=25\\ \Rightarrow x=42-25=17\)
a)x=2
b)x+8=14
x=6
c)\(42-x=5^2\)
\(42-x=25\)
\(-x=-17\)
\(x=17\)
\(a,\Leftrightarrow\left(x+8\right)\left(8x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=\dfrac{1}{8}\end{matrix}\right.\\ b,\Leftrightarrow\left(y-9\right)^2=0\Leftrightarrow y=9\)
a) ko có x ???
v) 27<813:3x>243
33<(34)3:3x>35
33<312:3x>35
33<312-x>35
=> 312-x=34
<=> 12-x=4
=>x=8
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