18-4x=6
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a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
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Bổ đề a^3+b^3+c^3-3abc= 0
<=> (a+b+c)[a^2+b^2+c^2-ab-bc-ca]=0
<=> 1/2 .(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]=0
<=> a+b+c=0 hoặc a=b=c
Đặt u =x^2-3 , v= - (4x+6 )
Ta có u^3+v^3 +216 = 18.u.v
<=> u^3+v^3+6^3 - 3.6.uv=0
Áp dụng bổ đề
=> u=v=3 hoặc u+v+3=0
*TH1: u=v=3 => x^2-3=3 và 4x+6=-3 ( vô lý)
*TH2 : u+v+3=0 <=> x^2-3-(4x+6)+3=0 <=> x^2-4x-6=0
=> x=2+√10 hay x=2-√10
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\(\sqrt{x+2}+\sqrt{9x+18}=\sqrt{4x+8+6}\)
\(\sqrt{x+2}+\sqrt{9\left(x+2\right)}=\sqrt{4x+14}\) \(\left(Đk:x\ge-2\right)\)
\(4\sqrt{x+2}=\sqrt{4x+14}\)
\(16\left(x+2\right)=4x+14\)
\(12x=-18\)
\(x=-\dfrac{3}{2}\left(TM\right)\)
Vậy \(S=\left\{-\dfrac{3}{2}\right\}\)
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18 - 4\(x\) = -20 - 6\(x\)
-4\(x\) + 6\(x\) = - 20 - 18
2\(x\) = - 38
\(x\) = - 19
h, -15 \(\times\) 24 = -7\(x\) + 32
7\(x\) = 360 + 32
7\(x\) = 392
\(x\) = 392:7
\(x\) = 56
i, 15\(x\) -3.(4\(x\) - 6) = -12 + 36
15\(x\) - 12\(x\) + 18 = 24
3\(x\) = 24 - 18
3\(x\) = 6
\(x\) = 2
k, -10\(x\) - 27 = -7\(x\) + 33
-27 - 33 = -7\(x\) + 10\(x\)
3\(x\) = -60
\(x\) = -20
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a, \(x+18\) = 6 - 2\(x\)
\(x\) + 2\(x\) = 18 - 6
3\(x\) = 12
\(x\) = 4
b, 17 - \(x\) = 7 - 6\(x\)
- \(x\) + 6\(x\) = 7 - 17
5\(x\) = - 10
\(x\) = - 2
c, \(x\) + 15 = 20 - 4\(x\)
\(x\) + 4\(x\) = 20 - 15
5\(x\) = 5
\(x\) = 1
d, 7\(x\) - 4 = 20 + 3\(x\)
7\(x-3x=20+4\)
4\(x\) = 24
\(x\) = 6
e, -12 + \(x\) = 5\(x\) - 20
-12 + 20 = 5\(x\) - \(x\)
4\(x\) = 8
\(x\) = 2
f, 4\(x\) - 18 = \(x\) + 27
4\(x\) - \(x\) = 27 + 18
3\(x\) = 45
\(x\) = 15
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`a, <=> 5/3 . 3sqrt(x^2+2) + 3/2.2sqrt(x^2+2)-7sqrt6=sqrt(x^2+2)`
`= (5+3-1)sqrt(x^2+2)=7sqrt6`
`<=> 7sqrt(x^2+2)=7sqrt6`.
`<=> x^2+2=36`.
`<=> x^2=34`.
`<=> x=+-sqrt(34)`.
Vậy...
`b, sqrt(4x^2-12x+9)-6=0`
`<=> |2x-3|=6`.
`@ x >=3/2 <=> 2x-3=6.`
`<=> x=9/2 (tm)`.
`@x <3/2 <=> 3-2x=6`
`<=> 2x=-3`
`<=> x=-3/2.`
Vậy...
\(18-4x=6\)
\(\Rightarrow4x=18-6\)
\(\Rightarrow4x=12\)
\(\Rightarrow x=3\)
\(18-4x=6\\ 4x=18-6\\ 4x=12\\ x=12:4\\ x=3.\)