a) \(9x^2-6x-3=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

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

b) \(x^3+9x^2+27x+19=0\)

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

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

\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))

c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)

\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)

a)\(9x^2-6x-3=0\)

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

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

\(\Leftrightarrow\)\((3x-1)(x-1)=0\)

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

        27x³-27x²+9x-1=1

<=>  (3x)³-3.9x².1+3.3.1²-1³ =1  (hằng đẳng thức thứ năm)

<=>  (3x-1)³=1

<=>  3x-1=1

<=> 3x=2

<=> x=2/3

Nhớ k đúng cho mik nha

27x³ - 27x² + 9x -1 =1

<=>(3x-1)³=1    (hằng đẳng thức thứ 5)

<=> 3x-1=1

<=> x=2/3

(mk tóm tắt bn tự hiểu nha)

a: Ta có: \(A=126y^3+\left(x-5y\right)\left(x^2+5xy+25y^2\right)\)

\(=126y^3+x^3-125y^3\)

\(=x^3+y^3\)

\(=\left(-5\right)^3+\left(-3\right)^3=-125-27=-152\)

b: \(C=x^3-9x^2+27x-26\)

\(=x^3-9x^2+27x-27+1\)

\(=\left(x-3\right)^3-1=\left(23-3\right)^3-1=20^3-1\)

=8000-1

=7999

c: \(D=\left(2x-3\right)^2+\left(4x-6\right)\left(4-x\right)+\left(x-4\right)^2\)

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

\(=\left(2x-3-x+4\right)^2=\left(x+1\right)^2\)

\(=\left(99+1\right)^2=100^2=10000\)

mày viết lại cái đề bài hộ tao cái

lm heets cmnr

\(x^3+9x^2+27x+26=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+7x+13\right)=0\Rightarrow x=-2\)

\(x^3+9x^2+27x+26=0\)

\(\Leftrightarrow x^3+9x^2+27x+27=1\)

\(\Leftrightarrow\left(x+3\right)^3=1^3\)

\(\Leftrightarrow x+3=1\Leftrightarrow x=-2\)