x^3+5x^2+8x+4=0

x^3+x^2+4x^2+4x+4x+4=0

x^2(x+1)+4x(x+1)+4(x+1)=0

(x+1)(x^2+4x+4)=0

x+1=0 =>x=-1

x^2+4x+4=0

x^2+2x+2x+4=0

x(x+2)+2(x+2)=0

(x+2)^2=0

x=-2

Vậy x=-2,x=-1

x^5 -x^3 -x^2 +1=0

x^3(x^2 -1 )-(x^2-1)=0

(x-1)(x^2+x+1)(x-1)(x+1)=0

(x-1)^2(x+1)(x^2+x+1)=0

=> x=1;x=-1

x^3- 5x^2+ 8x- 4= x^3- x^2- 4x^2+ 4x+ 4x- 4

                       = x^2(x-1)- 4x(x-1)+4(x-1)

                       = (x-1)(x^2-4x+4)

                       = (x-1)(x-1)^2

                       =(x-1)^3

1,

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

=> x=1 hoặc x=2

2, 

<=>\(\left(x+1\right)\left(2x^2-3x+6\right)\)=0

=> x=-1

1.

<=> ( x -1 ) ( x - 2 ) ² = 0

=> x = 1 hoặc x = 2

2.

<=> ( x + 1 ) ( 2x² - 3x + 6 ) = 0

=> x = -1

x=0

#Tự_tìm_lời_giải_nha_NGẠI_GHI

#Hk_tốt

#Ngọc's_Ken

\(5x^2+\left(-8x\right)=0\)

\(\Rightarrow5x^2-8x=0\)

\(\Rightarrow x\left(5x-8\right)=0\)

Từ đó ta xét 2 trường hợp :

TH1 :\(x=0\)

TH2 : \(5x-8=0\)

\(\Rightarrow5x=8\)\(\Rightarrow x=\frac{8}{5}\)

\(\Rightarrow x\in\left\{0;\frac{8}{5}\right\}\)

    5x² + (-8x) = 0

=>  5x² - 8x = 0

=>  x(5x - 8) =0

=>  \(\orbr{\begin{cases}x=0\\5x-8=0\end{cases}}\)

=>   \(\orbr{\begin{cases}x=0\\5x=8\end{cases}}\)

=>    \(\orbr{\begin{cases}x=0\\x=\frac{8}{5}\end{cases}}\)

Vậy x=0 hoặc x=\(\frac{8}{5}\)

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

a, 4x² - 49 = 0

⇔⇔ (2x)² - 7² = 0

⇔⇔ (2x - 7)(2x + 7) = 0

⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72

b, x² + 36 = 12x

⇔⇔ x² + 36 - 12x = 0

⇔⇔ x² - 2.x.6 + 6² = 0

⇔⇔ (x - 6)² = 0

⇔⇔ x = 6

e, (x - 2)² - 16 = 0

⇔⇔ (x - 2)² - 4² = 0

⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0

⇔⇔ (x - 6)(x + 2) = 0

⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2

f, x² - 5x -14 = 0

⇔⇔ x² + 2x - 7x -14 = 0

⇔⇔ x(x + 2) - 7(x + 2) = 0

⇔⇔ (x + 2)(x - 7) = 0

⇔{x+2=0x−7=0⇔{x=−2x=7

a,\(4x^2-49=0\)

\(\Leftrightarrow\left(2x\right)^2-7^2=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}}\)

b.\(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

c.\(\frac{1}{16x^2}-x+4=0\)

\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)

\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)

........