a) \(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

b) \(3x\left(x-2\right)-5x+10=0\)

\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)

c) \(4x\left(x+3\right)-x^2+9=0\)

\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)

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

\(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)

f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)

g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)

a) 4x³ - 9x = 0

<=> x( 4x² - 9 ) = 0

<=> x( 2x - 3 )( 2x + 3 ) = 0

<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x + 3 = 0

<=> x = 0 hoặc x = ±3/2

b) 3x( x - 2 ) - 5x + 10 = 0

<=> 3x( x - 2 ) - 5( x - 2 ) = 0

<=> ( x - 2 )( 3x - 5 ) = 0

<=> x - 2 = 0 hoặc 3x - 5 = 0

<=> x = 2 hoặc x = 5/3

c) 4x( x + 3 ) - x² + 9 = 0

<=> 4x( x + 3 ) - ( x² - 9 ) = 0

<=> 4x( x + 3 ) - ( x - 3 )( x + 3 ) = 0

<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0

<=> ( x + 3 )( 4x - x + 3 ) = 0

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

<=> x + 3 = 0 hoặc 3x + 3 = 0

<=> x = -3 hoặc x= -1

d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )

<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0

<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0

<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0

<=> ( x - 4 ).3x = 0

<=> x - 4 = 0 hoặc 3x = 0

<=> x = 4 hoặc x = 0

e) 16x² - 25 = ( 4x - 5 )( 2x + 1 )

<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0

<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0

<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0

<=> ( 4x - 5 )( 2x + 4 ) = 0

<=> 4x - 5 = 0 hoặc 2x + 4 = 0

<=> x = 5/4 hoặc x = -2

f) ( x + 1/5 )² = 64/9

<=> ( x + 1/5 )² = ( ±8/3 )²

<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3

<=> x = 37/15 hoặc x = -43/15

g) 9( x + 2 )² = ( x + 3 )²

<=> 3²( x + 2 )² - ( x + 3 )² = 0

<=> [ 3( x + 2 ) ]² - ( x + 3 )² = 0

<=> ( 3x + 6 )² - ( x + 3 )² = 0

<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0

<=> ( 3x + 6 - x - 3 )( 3x + 6 + x + 3 ) = 0

<=> ( 2x + 3 )( 4x + 9 ) = 0

<=> 2x + 3 = 0 hoặc 4x + 9 = 0

<=> x = -3/2 hoặc x = -9/4

a) p² - 2pq + q² - 9 ( sửa rồi nhé :)) )

= ( p² - 2pq + q² ) - 9

= ( p - q )² - 3²

= ( p - q - 3 )( p - q + 3 )

b) ( a + 3 )² + ( a² - 9 )² ( sửa rồi nhé p2 :)) )

= ( a + 3 )² + [ ( a - 3 )( a + 3 ) ]²

= ( a + 3 )² + ( a - 3 )²( a + 3 )²

= ( a + 3 )²[ 1 + ( a - 3 )² ]

= ( a + 3 )²( 1 + a² - 6a + 9 )

= ( a + 3 )²( a² - 6a + 10 )

a) Ta có: \(\left(x-3\right)\left(1-x\right)-2\)

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

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

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

=> đpcm

b) Ta có: \(\left(x+4\right)\left(2-x\right)-10\)

\(=-x^2-2x+8-10\)

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

\(=-\left(x+1\right)^2-1\le-1< 0\left(\forall x\right)\)

=> đpcm

b) \(\left(x+4\right)\left(2-x\right)-10=-x^2-4x+2x+8-10\)

\(=-x^2-2x-2=-x^2-2x-1-1\)

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

Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+2\right)^2-1\le-1\forall x\)

\(\Rightarrow\left(x+4\right)\left(2-x\right)-10\le-1\forall x\)

hay \(\left(x+4\right)\left(2-x\right)-10\)luôn âm với mọi x ( đpcm )

Phân tích thôi hả :v 

( 3x - 7 )² - 25 = ( 3x - 7 )² - 5² = ( 3x - 7 - 5 )( 3x - 7 + 5 ) = ( 3x - 12 )( 3x - 2 ) = 3( x - 4 )( 3x - 2 )

( x - 1/2 )² - 9/4 = ( x - 1/2 )² - ( 3/2 )² = ( x - 1/2 - 3/2 )( x - 1/2 + 3/2 ) = ( x - 2 )( x + 1 )

49 - ( x + 7 )² = 7² - ( x + 7 )² = [ 7 - ( x + 7 ) ][ 7 + ( x + 7 ) ] = ( 7 - x - 7 )( 7 + x + 7 ) = -x( x + 14 )

25 - ( x - 3 )² = 5² - ( x - 3 )² = [ 5 - ( x - 3 ) ][ 5 + ( x - 3 ) ] = ( 5 - x + 3 )( 5 + x - 3 ) = ( 8 - x )( x + 2 )

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

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

\(X=1\)

( 2x + 1 )² = ( x + 2 )²

<=> ( 2x + 1 )² - ( x + 2 )² = 0

<=> [ ( 2x + 1 ) - ( x + 2 ) ][ ( 2x + 1 ) + ( x + 2 ) ] = 0

<=> ( 2x + 1 - x - 2 )( 2x + 1 + x + 2 ) = 0

<=> ( x - 1 )( 3x + 3 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\3x+3=0\end{cases}}\Leftrightarrow x=\pm1\)

\(\frac{1}{9}x^2-\frac{1}{16}=\left(\frac{1}{3}x\right)^2-\left(\frac{1}{4}\right)^2=\left(\frac{1}{3}x-\frac{1}{4}\right)\left(\frac{1}{3}x+\frac{1}{4}\right)\)

\(2-x^2=\left(\sqrt{2}\right)^2-x^2=\left(\sqrt{2}-x\right)\left(\sqrt{2}+x\right)\)

\(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

\(5-x^2=\left(\sqrt{5}\right)^2-x^2=\left(\sqrt{5}-x\right)\left(\sqrt{5}+x\right)\)

\(\frac{1}{4}x^4-9\)

\(=\left(\frac{1}{2}x^2\right)^2-3^2\)

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

HAHA :) 4 năm vẫn chưa có ai trả lời, nhục !