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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 = \(\frac{1}{3}\)
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
a/ (x-a)^4 - (x+a)^4
=((x-a)^2)^2 - ((x+a)^2)^2
=(x^2 - 2xa + a^2)^2 - (x^2 +2xa+a^2)^2
=(x^2-2xa+a^2-x^2-2xa-a^2)(x^2-2xa+a^2+x^2+2xa+a^2)
=-4xa(2x^2+2a^2)
b/ x^4 –y^2(2x-y)^2
=(x^2)^2-(y(2x-y)^2
=(x^2)^2-(2xy-y^2)^2
=(x^2-2xy+y^2)(x^2+2xy+y^2)
=(x-y)^2 (x+y)^2
c/(xy+4)^2- 4(x+y)^2
=(xy+4)^2- (2x+2y)^2
=(xy+y-2x-2y)(xy+y+2x+2y)
=(xy-y+2x)(xy+3y+2x)
a) 4x2-1=(2x)2-12=(2x-1)(2x+1)
b)25x2-0.09=(5x)2-\(\left(\frac{3}{10}\right)^2\)=\(\left(5x-\frac{3}{10}\right)\left(5x+\frac{3}{10}\right)\)
c)9x2-14=(3x)2-\(\sqrt{14}^2\)=(3x-\(\sqrt{14}\))(3x+\(\sqrt{14}\))
d) (x-y)2-4=(x-y)2-22=(x-y-2)(x-y+2)
e) 9-(x-y)2=33-(x-y)2=(3-x+y)(3+x-y)
f)(x2+4)2-16x2=(x2+4)2-(4x)2=(x2+4+4x)(x2+4-4x)
Chúc hok tốt!!!
\(a,\)\(4x^2-1\)\(=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
\(b,\)\(25x^2-0,09=\left(5x\right)^2-0,3^2=\left(5x-0,3\right)\left(5x+0,3\right)\)
\(c,\)\(9x^2-14=\left(3x\right)^2-\left(\sqrt{14}\right)^2=\left(3x-\sqrt{14}\right)\left(3x+\sqrt{14}\right)\)
\(d,\)\(\left(x-y\right)^2-4=\left(x-y\right)^2-2^2=\left(x-y+2\right)\left(x-y-2\right)\)
\(e,\)\(9-\left(x-y\right)^2=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
\(f,\)\(\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)^2-\left(4x\right)^2=\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
\(=\left(x^2-2.2x+2^2\right)\left(x^2+2.2x+2^2\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)
\(=3\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(3+2x-y\right)\left(2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)\)
\(=9x^2-\left(y-2\right)^2\)
\(=\left(3x+y-2\right)\left(3x-y+2\right)\)
A = ( 6x - 3y ) + (4x2 - 4xy + y2 )
A = 3.( 2x - y) + [ ( 2x )2 - 2.2.x.y + y2 ]
A = 3.( 2x - y ) + ( 2x - y )2
A = ( 2x - y ).(3 + 2x - y )
B = 9x2 - ( y2 - 4y + 4 )
B = ( 3x )2 - ( y - 2 )2
B = ( 3x - y + 2 ).( 3x + y - 2 )
C = - 25x2 + y2 - 6y + 9
C = ( y2 - 2.3.y + 32 ) - ( 5x )2
C = ( y - 3 )2 - ( 5x )2
C = (y - 3 - 5x ).( y - 3 +5x )
D = x2 - 4x - y2 -- 8y - 12
D = ( x2 - 4x + 4 ) - 4 - y2 - 8y -12
D = ( x - 2.2x + 22 ) - ( y2 + 2.4.y + 42 )
D = ( x - 2 )2 - ( y + 4 )2
D = ( x - 2 + y + 4 ).( x - 2 - y - 4 )
D = ( x + y + 2 ).( x - y - 6 )