\(\Leftrightarrow x^4+x^3-10x^2+1=x^3-8\)

\(\Leftrightarrow x^4-10x^2+9=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\\x=-3\end{matrix}\right.\)

a: x^2=16

=>x=4 hoặc x=-4

b: =>x^2=144

=>x=12 hoặc x=-12

a)x^2 =16

  x =4^2

b)x^2+(-4) = 140

  x^2          = 140-(-4) 

  x^2          = 144

  x^2          =12^2

2(x⁴+3)-(9)=17

⇒2x⁴+6+9=17

⇒2x⁴+15=17

⇒ 2x⁴=2

⇒ x⁴=1

⇒ x=\(\pm1\)

b) 5x².x+1-3.4²=-47

⇒5x³+1-48=-47

⇒5x³-47=-47

⇒5x³=0

⇒x³=0

⇒x=0

a) \(2\left(x^4+3\right)-\left(-9\right)=17\)

              \(2x^4+6+9=17\)

                           \(2x^4=2\)

                             \(x^4=1\)

                       ⇒     \(x=1\)

a) 2*(x⁴+3)-(-9)=17

=>2x⁴+2*3+9=17

=>2x⁴+6=17-9=8

=>2x⁴=8-6=2

=>x⁴=2/2=1

=>x=1

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

a ,\(4x^2-\left(x-3\right)^2=0\)

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

\(\Leftrightarrow\left(x+3\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\3x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Vậy 

b,\(x^2-4+\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)^2=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy ...

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)

\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)

\(x^2+6x+9-x^2+9=0\)

\(6x+18=0\)

\(6x=-18\)

\(x=-3\)

Vậy x=-3

\(b,5x^3+20x=0\)

\(5x\left(x^2+4\right)=0\)

\(Th1:5x=0=>x=0\)

\(Th2:x^2+4=0\)

\(x^2=-4\)(vô lý)

Vậy x=0

a) \(\Rightarrow3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

b) \(\Rightarrow x^3+6x^2+12x+8-x^3+6x^2=4\)

\(\Rightarrow12x^2+12x+4=0\)

\(\Rightarrow x\in\varnothing\)(do \(12x^2+12x+4=12\left(x^2+x+\dfrac{1}{4}\right)+1=12\left(x+\dfrac{1}{2}\right)^2+1\ge1>0\))