3^8x+4=81^x=3
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\(3\sqrt[3]{x-3}+4\sqrt[3]{8x-24}-\frac{1}{3}\sqrt[3]{27x-81}=3\sqrt[3]{x-3}+4\sqrt[3]{8\left(x-3\right)}-\frac{1}{3}\sqrt[3]{27\left(x-3\right)}=3\sqrt[3]{x-3}+4.2.\sqrt[3]{x-3}-\frac{1}{3}.3.\sqrt[3]{x-3}=3\sqrt[3]{x-3}+8\sqrt[3]{x-3}-\sqrt[3]{x-3}=10.\sqrt[3]{x-3}=-20\Leftrightarrow\sqrt[3]{x-3}=-2\Leftrightarrow x-3=-8\Leftrightarrow x=-5\)
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
a) \(x^4-4x^{3^{ }}+8x+3\)
\(=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(5x^2+5x\right)+\left(3x+3\right)\)
\(=x^{3^{ }}\left(x+1\right)-5x^{2^{ }}\left(x+1\right)+5x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-5x^2+5x+3\right)\)
\(=\left(x+1\right)\left[\left(x^3-3x^2\right)-\left(2x^2-6x\right)-\left(x-3\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x-3\right)-2x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x^2-2x-1\right)\)
\(=\left(x+1\right)\left(x-3\right)\left[\left(x-1\right)^2-2\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)\)
b, \(x^2\left(y^2-4\right)^2-6x\left(y^2-4\right)+9\)
\(=\left[x\left(y^2-4\right)-3\right]^2\)
\(=\left(xy^2-4x-3\right)^2\)
b: 1/2x-4=0
=>1/2x=4
hay x=8
a: x+7=0
=>x=-7
e: 4x2-81=0
=>(2x-9)(2x+9)=0
=>x=9/2 hoặc x=-9/2
g: x2-9x=0
=>x(x-9)=0
=>x=0 hoặc x=9
a: x+7=0
nên x=-7
b: x-4=0
nên x=4
c: -8x+20=0
=>-8x=-20
hay x=5/2
d: x2-100=0
=>(x-10)(x+10)=0
=>x=10 hoặc x=-10
Lời giải:
$3^{8x+4}=81^{x+3}$
$3^{4(2x+1)}=(3^4)^{x+3}$
$3^{4(2x+1)}=3^{4(x+3)}$
$\Rightarrow 4(2x+1)=4(x+3)$
$\Rightarrow 2x+1=x+3$
$\Leftrightarrow x=2$ (tm)