(2x+3).(10x+2)=(5x+2).(4x+5)

b) 9x^2-(x-4)^2+0

=>x-3=0

hay x=3

\(\left(x-3\right)\left(2x^2+3\right)=0\\ \Rightarrow x-3=0\left(vì.2x^2+3>0\right)\\ \Rightarrow x=3\)

( x − 3 ) ( 2 x 2 + 3) = 0

⇒ x − 3 = 0 ( vì . 2 x 2 + 3 > 0 )

⇒ x = 3

\(\Leftrightarrow x\in\left\{0;43\right\}\)

\(x+x+x\times1+x\times6=120\\ x\times\left(1+1+1+6\right)=120\\ x\times9=120\\ x=\dfrac{120}{9}\\ x=\dfrac{40}{3}\)

\(x+x+x\text{×}1+x\text{×}6=120\)

\(x\text{×}\left(1+1+1\right)+x\text{×}6=120\)

\(x\text{×}3+x\text{×}6=120\)

\(x\text{×}\left(3+6\right)=120\)

\(x\text{×}9=120\)

\(x=\dfrac{120}{9}\)

x+x+x×1+x×6=120x×(1+1+1+6)=120x×9=120x=1209x=403

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^2-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x\left(x-2\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-3;0;2\right\}\)

\(x\left(x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x=0\) hay \(x+3=0\) hay \(x^2+1=0\) (pt vô nghiệm vì \(x^2+1\ge1\))

\(\Leftrightarrow x=0\) hay \(x=-3\)

-Vậy \(S= \left\{0;-3\right\}\)

\(a,2^{x+1}=64\\ \Rightarrow a,2^{x+1}=2^6\\ \Rightarrow x+1=6\\ \Rightarrow x=5\)

\(b,x=18\)

\(c,\left(4x-9\right)-\left(x+111\right)=0\\ \Rightarrow4x-9-x-111=0\\ \Rightarrow3x-120=0\\ \Rightarrow3x=120\\ \Rightarrow x=40\)

a
,
2
x
+
1
=
64
⇒
a
,
2
x
+
1
=
2
6
⇒
x
+
1
=
6
⇒
x
=
5

b
,
x
=
18

c
,
(
4
x
−
9
)
−
(
x
+
111
)
=
0
⇒
4
x
−
9
−
x
−
111
=
0
⇒
3
x
−
120
=
0
⇒
3
x
=
120
⇒
x
=
40