a, \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

b, \(27x^3+27x^2+9x+1=0\Leftrightarrow27x^3+1+27x^2+9x=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2-3x+1\right)+9x\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2+2>0\right)=0\Leftrightarrow x=-\frac{1}{3}\)

c, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=\frac{2}{3};x=-1\)

d, \(\left(x+1\right)^3-25\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\Leftrightarrow x=-1;x=-6;x=4\)

a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)

Ta có: \(x^2-4x+5\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

\(\Leftrightarrow x\left(x+3\right)-5\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

Vậy: x∈{-3;5}

d) Ta có: \(\left(x^2-1\right)^2=4x+1\)

\(\Leftrightarrow x^4-2x^2+1-4x-1=0\)

\(\Leftrightarrow x^4-2x^2-4x=0\)

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

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

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

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

Ta có: \(x^2+2x+2\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

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

Vậy: x∈{0;2}

a) 5x + 6 = 0

<=> 5x = -6

<=> x = \(-\frac{6}{5}\)

Vậy phương trình có tập nghiệm là: S = {\(-\frac{6}{5}\)}
b) 9x - 3 = 6x + 21

<=> 3x = 24

<=> x = 8

Vậy phương trình có tập nghiệm là: S = {8}
c) x³ - 9x = 0

<=> x(x² - 9) = 0

<=> x(x - 3)(x + 3) = 0

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

Vậy phương trình có tập nghiệm là: S = {0; 3; -3}
d) ĐKXĐ: \(x\ne2;x\ne-2\)

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

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

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(TM\right)\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: S ={1}

a) Ta có: 5x+6=0

⇔5x=-6

hay \(x=-\frac{6}{5}\)

Vậy: \(S=\left\{-\frac{6}{5}\right\}\)

b) Ta có: 9x-3=6x+21

⇔9x-6x=21+3

⇔3x=24

hay x=8

Vậy: S={8}

c) Ta có: \(x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

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

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

Vậy: S={-3;0;3}

d) ĐKXĐ: x∉{2;-2}

Ta có: \(\frac{1}{x-2}-\frac{x^2-4}{4-x^2}=0\)

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

\(\Leftrightarrow\frac{1}{x-2}+1=0\)

\(\Leftrightarrow\frac{1}{x-2}+\frac{x-2}{x-2}=0\)

Suy ra: \(1+x-2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(tm)

Vậy: S={1}

hết cứu đi mà làm

a) \(x^3+9x^2+27x+19=0\)

\(\Rightarrow x^3+x^2+8x^2+8x+19x+19=0\)

\(\Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\)

Mà \(x^2+8x+19=x^2+2.x.4+16+3=\left(x+4\right)^2+3\)

Vì \(\left(x+4\right)^2\ge0\) với mọi x

\(3>0\)

\(\Rightarrow\left(x+4\right)^2+3>0\) với mọi x

=> ( x + 4 )² + 3 vô nghiệm

=> x + 1 = 0

=> x = -1

Vậy x = -1

b) \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)

\(\Rightarrow\left(2x\right)^3+3.\left(2x\right)^2+3.2x+1+x\left(x^2-2^2\right)-9x\left(x^2-4x+4\right)+57=0\)

\(\Rightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)

\(\Rightarrow48x^2-34x+58=0\)

\(\Rightarrow2\left(24x^2-17x+29\right)=0\)

\(\Rightarrow24x^2-17x+29=0\)

... Tới đây mình bí luôn rồi, sorry

Câu a : \(x^3+9x^2+27x+19=0\)

\(\Leftrightarrow\left(x^3+9x^2+27x+27\right)-8=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Leftrightarrow x+1=0\) ( Vì : \(x^2+8x+19>0\))

\(\Leftrightarrow x=-1\)

Vậy \(x=-1\)

Câu b : \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)

\(\Leftrightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)

\(\Leftrightarrow48x^2-34x+58=0\)

\(\Rightarrow PTVN\)

Vậy ko có giá trị của x

a)Ta có: x^3 +9x^2 +27x + 19=0

x^3 + 9x^2 + 27x + 27 -8=0

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

(x+3-2)(x^2 + 6x+9 +4x+12 +4)=0

(x+1)(x^2 +10x+ 25)=0

=> x+1=0 hoặc x^2 +10x+25=0

=> x=-1 hoặc (x+5)^2 =0

=> x=-1 hoặc x+5=0

=> x=-1 hoặc x=-5

Em ơi mình đăng bài sang bên môn toán nha

( x - 2 ) ³ - ( x + 1 ) ³ + 9x ( x + 1 ) - 9 = 0
=> \(x^3-6x^2+12x-8-\left(x^3+3x^2+3x+1\right)+9x^2+9x-9=0\)
=> \(x^3-6x^2+12x-8-x^3-3x^2-3x-1+9x^2+9x-9=0\)
=> \(18x-18=0\)
=>         \(18x=0+18\)
=>         \(18x=18\)
=>             \(x=1\)