a: \(x^3-4x^2-x+4=0\)

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

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

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a) $(x-3)^2-(x+2)(x-2)=-5$

$\Rightarrow x^2-2\cdot x\cdot3+3^2-(x^2-2^2)=-5$

$\Rightarrow x^2-6x+9-(x^2-4)=-5$

$\Rightarrow x^2-6x+9-x^2+4=-5$

$\Rightarrow-6x+13=-5$

$\Rightarrow-6x=-18$

$\Rightarrow x=3$

b) $x^3-2x^2-4x+8=0$

$\Rightarrow(x^3-2x^2)-(4x-8)=0$

$\Rightarrow x^2(x-2)-4(x-2)=0$

$\Rightarrow (x^2-4)(x-2)=0$

$\Rightarrow (x^2-2^2)(x-2)=0$

$\Rightarrow (x-2)(x+2)(x-2)=0$

$\Rightarrow (x-2)^2(x+2)=0$

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

$\text{#}Toru$

sao ko có = máy hít vậy

a)8x²+30x+7=0

=>8x²+28x+2x+7=0

=>(8x²+2x)+(28x+7)=0

=>2x(4x+1)+7(4x+1)=0

=>(2x+7)(4x+1)=0

\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)

b)(x²-4x)²-8(x²-4x)+15=0

=>x⁴-8x³+8x²+32x+15=0

=>(x-5)(x+1)(x²-4x-3)=0

\(\Rightarrow\hept{\begin{cases}x=5\\x=-1\\x=2-\sqrt{7};x=\sqrt{7}+2\end{cases}}\)

x²+5x+6
 

\(a,3\sqrt{x}-7=0\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=7\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{3}\\ \Leftrightarrow x=\dfrac{49}{9}\left(tmdk\right)\)

Vậy \(S=\left\{\dfrac{49}{9}\right\}\)

\(b,\sqrt{x-2}+\sqrt{4x-8}=3\left(dk:x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}=3\\ \Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}=3\\ \Leftrightarrow3\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tmdk\right)\)

Vậy \(S=\left\{3\right\}\)

a: =>3*căn x=7

=>căn x=7/3

=>x=49/9

b: =>3*căn x-2=3

=>căn x-2=1

=>x-2=1

=>x=3

`a, 3 sqrt x - 7 = 0`

`<=> 3 sqrt x = 7`

`<=> sqrt x = 7/3`

`<=> x = 49/9`.

Vậy `x = 49/9`

`b, sqrt(x-2) + sqrt(4x-8) = 3`

`<=> sqrt(x-2) + 2 sqrt(x-2) = 3`

`<=> 3 sqrt(x-2) = 3`

`<=> sqrt(x-2)=1`

`<=> x-2=1`

`<=> x = 3`

Vậy `x = 3`

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)