1)3.x^2 - 75 = 0

3.x^2 - 3.25 = 0

3.(x^2-25)=0

x^2-5^2=0

(x-5)(x+5)=0

=> x-5=0 hoặc x+5=0

=> x=5 hoặc x=-5

1) \(3x^2-75=0\)

\(\Leftrightarrow3\left(x^2-25\right)=0\)

\(\Leftrightarrow x^2-25=0\)

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)

2) \(x^3+9x^2+27x+27=0\)

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

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

3) \(x^3+3x^2+3x=0\)

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

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

\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)

2, x^3 + 9x^2 +27x + 27 = 0

x^3 + 3.x^2.3 + 3.x.3^2 +3^3 = 0

(x+3)^3=0

x+3=0

x=-3

1/ \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

\(\Leftrightarrow9x^2-6x-35=0\)

\(\Leftrightarrow\left(2x-1\right)^2-36=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+6\right)=0\)

2/ \(\left(3x+5\right)^2-4x^2=0\)

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

3/ \(25x^2-\left(4x-3\right)^2=0\)

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

1) ( 9x² - 25 ) - ( 6x - 10 ) = 0

\(\Leftrightarrow\) [ ( 3x)² - 5² ] - 2.( 3x + 5 ) = 0

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

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

\(\Leftrightarrow\)( 3x + 5 ).( 3x + 3 ) = 0

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+5=0\\3x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-5\\3x=-3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{-5}{3}\\x=-1\end{cases}}\)

Vậy x = \(\frac{-5}{3}\) , x = -1

2) ( 3x + 5 )² - 4x²  = 0

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

\(\Leftrightarrow\)( x + 5 ).( 5x + 5 ) = 0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+5=0\\5x+5=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-5\\x=-1\end{cases}}\)

Vậy x = -5 , x = -1

3) 25x² - ( 4x - 3 )² = 0

\(\Leftrightarrow\)( 5x )2 - ( 4x - 3 )² = 0

\(\Leftrightarrow\) ( 5x - 4x + 3 ).(5x + 4x - 3 ) = 0

\(\Leftrightarrow\)( x + 3 ).( 9x - 3 ) = 0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\9x-3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\9x=3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

Vậy x = 3 , x = \(\frac{1}{3}\)

a) =( x²+y²-5)²-[2(xy-2)]²

⁼( x²+y²-5)²- (2xy-4)²

=(x²+y²-5+2xy-4)(x²+y²-5-2xy+4)

=[(x+y)²-9][(x-y)²-1]

phân tích tiếp HĐT 2 ở 2 thừa số

bài 2 :

0,25x³+x²+x=0

<=>0,25x³+0,5x²+0,5x²+x=0

<=>0,25x²(x+2)+0,5x(x+2)=0

<=>(x+2)(0,25x²+0,5x)=0

<=>(x+2)x(0,25x+0,5)=0

<=>x+2=0 hoặc x=0 hoặc 0,25x+0,5=0

=>x=-2 hoặc x=0 hoặc x=-2

vậy x=0 hoặc x=-2

Lời giải:
$x^4y^4-z^4=(x^2y^2)^2-(z^2)^2=(x^2y^2-z^2)(x^2y^2+z^2)$

$=(xy-z)(xy+z)(x^2y^2+z^2)$

$(x+y+z)^2-4z^2=(x+y+z)^2-(2z)^2=(x+y+z-2z)(x+y+z+2z)$
$=(x+y-z)(x+y+3z)$

$\frac{-1}{9}x^2+\frac{1}{3}xy-\frac{1}{4}y^2=\frac{-4x^2+12xy-9y^2}{36}$

$=-\frac{4x^2-12xy+9y^2}{36}=-\frac{(2x-3y)^2}{36}=-\left(\frac{2x-3y}{6}\right)^2$

Câu trả lời của cô quá đúng luôn đấy

a) Ta có: \(x^4y^4-z^4\)

\(=\left(x^2y^2-z^2\right)\left(x^2y^2+z^2\right)\)

\(=\left(xy-z\right)\left(xy+z\right)\left(x^2y^2+z^2\right)\)

b) Ta có: \(\left(x+y+z\right)^2-4z^2\)

\(=\left(x+y+z-2z\right)\left(x+y+z+2z\right)\)

\(=\left(x+y-z\right)\left(x+y+3z\right)\)

(1)\(\left(3x\right)^2+2×3x+y^2=\left(3x+y\right)^2\)

(2)\(-x^2+6x-9=-\left(x^2-6x+9\right)=-\left(x^2-6x+3^2\right)=-\left[\left(x-3\right)^2\right]\)

(3)\(x^2+4xy+4y^2=x^2+2×x×2y^2+\left(2y\right)^2=\left(x+2y\right)^2\)

\(=\left(4-a+b\right)\left(4+a-b\right)\)

x^2 -3^2=0

(x+3).(x-3)=0

x+3=0 hoặc x-3=0

x=-3 hoặc x=3

        \(x^2-9=0\)

⇔ \(x^2-3^2=0\)

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

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

Vậy \(x=\pm3\)