(4x-2)6=64
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\(\Leftrightarrow\sqrt{x-4}\left(4-12\cdot\dfrac{1}{2}+2\cdot2\right)=6\)
=>x-4=9
hay x=13
\(318-5\left(x-64\right)=103\)
\(\Rightarrow5\left(x-64\right)=318-103\)
\(\Rightarrow5\left(x-64\right)=215\)
\(\Rightarrow x-64=43\)
\(\Rightarrow x=43+64\)
\(\Rightarrow x=107\)
_____________
\(4^x\cdot5+216=296\)
\(\Rightarrow4^x\cdot5=296-216\)
\(\Rightarrow4^x\cdot5-80\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
___________
\(376-6^x:3=364\)
\(\Rightarrow6^x:3=376-364\)
\(\Rightarrow6^x:3=12\)
\(\Rightarrow6^x=36\)
\(\Rightarrow6^x=6^2\)
\(\Rightarrow x=2\)
___________
\(\left(4x-1\right)^2=121\)
\(\Rightarrow\left(4x-1\right)^2=11^2\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=11\\4x-1=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=12\\4x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a.2x - 138 =72
\(\Leftrightarrow\)2x = 210 \(\Leftrightarrow\)x= 105
b.32 - 3x =66 : 65
\(\Leftrightarrow\)9 - 3x =6 \(\Leftrightarrow\)3x =3 \(\Leftrightarrow\)x=1
c.( 64 -4x).3 =24
\(\Leftrightarrow\)64-4x =8 \(\Leftrightarrow\)4x = 56 \(\Leftrightarrow\)x=14
d.3x.6 - 39 =123
\(\Leftrightarrow\) 3x . 6 = 162 \(\Leftrightarrow\)3x =27 \(\Leftrightarrow\)x =3
1: \(=x^4-8x^3+24x^2-32x+16+x^4-12x^3+54x^2-108x+81-1\)
\(=2x^4-20x^3+78x^2-140x+96\)
\(=2\left(x-3\right)\left(x-2\right)\left(x^2-5x+8\right)\)
2: \(=x^4-4x^3+6x^2-4x+1+x^4+12x^3+54x^2+108x+81-512\)
\(=2x^4+8x^3+60x^2+104x-430\)
\(=2\left(x^4+4x^3+30x^2+52x-215\right)\)
1.
$\sqrt{3x^2}-\sqrt{12}=0$
$\Leftrightarrow \sqrt{3x^2}=\sqrt{12}$
$\Leftrightarrow 3x^2=12$
$\Leftrightarrow x^2=4$
$\Leftrightarrow (x-2)(x+2)=0\Leftrightarrow x=\pm 2$
2.
$\sqrt{(x-3)^2}=9$
$\Leftrightarrow |x-3|=9$
$\Leftrightarrow x-3=9$ hoặc $x-3=-9$
$\Leftrightarrow x=12$ hoặc $x=-6$
a, \(49x^2-70x+25=\left(7x\right)^2-2.7x.5+5^2=\left(7x-5\right)^2\)
Thay x = 5 vào biểu thức trên : \(\left(35-5\right)^2=30^2=900\)
b, \(x^3+12x^2+48x+64=\left(x+4\right)^3\)
Thay x = 6 vào biểu thức trên ta được : \(\left(6+4\right)^3=1000000\)
3, \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Thay x = -6 ; y = 2 vào biểu thức trên ta được : \(\left(-12+2\right)^2=100\)
a) \(x^2-64=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
b) \(4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c) \(9-6x+x^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
a: Ta có: \(x^2-64=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
b: Ta có: \(4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
hay \(x=\dfrac{1}{2}\)
c: ta có: \(x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
hay x=3
(4x-2)6=64
= (4x-2)6 = 26
= 4 x - 2 = 2
= 4 x = 2 + 2
= 4 x = 4
= x = 4 : 4
= x = 1
`(4x-2)^6 = 64`
`(4x-2)^6 = `\(\left(\pm2\right)^6\)
`@TH1:`
`4x-2=2`
`4x=2+2`
`4x=4`
`x=4:4`
`x=1`
`@TH2:`
`4x-2=-2`
`4x=-2+2`
`4x=0`
`x=0:4`
`x=0`
Vậy `x = {0;1}`