Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
a. Biểu thức ko thể biểu diễn dưới dạng tích của các thừa số
b. (x-1)(4x+1)
c. -(3z^2-5y^2-6xy-3x^2)
d. x(y^2-2xy+x-9)
e. -(y-x)(y-x+2)
f. y^3+xy^2+3x^2y-y+x^2-x
HỌC TỐT.
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
a) ( 3x - 1 )2 - 16 = ( 3x - 1 )2 - 42 = ( 3x - 1 - 4 )( 3x - 1 + 4 ) = ( 3x - 5 )( 3x + 3 ) = 3( 3x - 5 )( x + 1 )
b) ( 5x - 4 )2 - 49x2 = ( 5x - 4 )2 - ( 7x )2 = ( 5x - 4 - 7x )( 5x - 4 + 7x ) = ( -2x - 4 )( 12x - 4 ) = -2( x + 2 ).4( 3x - 1 ) = -8( x + 2 )( 3x - 1 )
c) ( 2x + 5 )2 - ( x - 9 )2 = [ ( 2x + 5 ) - ( x - 9 ) ][ ( 2x + 5 ) + ( x - 9 ) ] = ( 2x + 5 - x + 9 )( 2x + 5 + x - 9 ) = ( x + 14 )( 3x - 4 )
d) ( 3x + 1 )2 - 4( x - 2 )2 = ( 3x + 1 )2 - 22( x - 2 )2 = ( 3x + 1 )2 - [ 2( x - 2 ) ]2 = ( 3x + 1 )2 - ( 2x - 4 )2 = [ ( 3x + 1 ) - ( 2x - 4 ) ][ ( 3x + 1 ) + ( 2x - 4 ) ] = ( 3x + 1 - 2x + 4 )( 3x + 1 + 2x - 4 ) = ( x + 5 )( 5x - 3 )
e) 9( 2x + 3 )2 - 4( x + 1 )2 = 32( 2x + 3 )2 - 22( x + 1 )2 = [ 3( 2x + 3 ) ]2 - [ 2( x + 1 ) ]2 = ( 6x + 9 )2 - ( 2x + 2 )2 = [ ( 6x + 9 ) - ( 2x + 2 ) ][ ( 6x + 9 ) + ( 2x + 2 ) ] = ( 6x + 9 - 2x - 2 )( 6x + 9 + 2x + 2 ) = ( 4x + 7 )( 8x + 11 )
f) 4b2c2 - ( b2 + c2 - a2 )2 = ( 2bc )2 - ( b2 + c2 - a2 )2 = [ 2bc - ( b2 + c2 - a2 ) ][ 2bc + ( b2 + c2 - a2 ] = ( 2bc - b2 - c2 + a2 )( 2bc + b2+ c2 - a2 ) = [ a2 - ( b2 - 2bc + c2 ) ][ ( b2 + 2bc + c2 ) - a2 ] = [ a2 - ( b - c )2 ][ ( b + c )2 - a2 ] = ( a - b + c )( a + b - c )( b + c - a )( b + c + a )
g) ( ax + by )2 - ( ay + bx )2
= [ ( ax + by ) - ( ay + bx ) ][ ( ax + by ) + ( ay + bx ) ]
= ( ax + by - ay - bx )( ax + by + ay + bx )
= [ a( x - y ) - b( x - y ) ][ a( x + y ) + b( x + y ) ]
= ( a - b )( x - y )( x + y )( a + b )
h) ( a2 + b2 - 5 )2 - 4( ab + 2 )2
= ( a2 + b2 - 5 )2 - 22( ab + 2 )2
= ( a2 + b2 - 5 )2 - [ 2( ab + 2 ) ]2
= ( a2 + b2 - 5 )2 - ( 2ab + 4 )2
= [ ( a2 + b2 - 5 ) - ( 2ab + 4 ) ][ ( a2 + b2 - 5 ) + ( 2ab + 4 ) ]
= ( a2 + b2 - 5 - 2ab - 4 )( a2 + b2 - 5 + 2ab + 4 )
= [ ( a2 - 2ab + b2 ) - 9 ][ ( a2 + 2ab + b2 ) - 1 ]
= [ ( a - b )2 - 32 ][ ( a + b )2 - 12 ]
= ( a - b - 3 )( a - b + 3 )( a + b - 1 )( a + b + 1 )
i) ( 4x2 - 3x - 18 )2 - ( 4x2 + 3x )2
= [ ( 4x2 - 3x - 18 ) - ( 4x2 + 3x ) ][ ( 4x2 - 3x - 18 ) + ( 4x2 + 3x ) ]
= ( 4x2 - 3x - 18 - 4x2 - 3x )( 4x2 - 3x - 18 + 4x2 + 3x )
= ( -6x - 18 )( 8x2 - 18 )
= -6( x + 3 ).2( 4x2 - 9 )
= -12( x + 3 )( 2x - 3 )( 2x + 3 )
k) 9( x + y - 1 )2 - 4( 2x + 3y + 1 )2
= 32( x + y - 1 )2 - 22( 2x + 3y + 1 )2
= [ 3( x + y - 1 ) ]2 - [ 2( 2x + 3y + 1 ) ]2
= ( 3x + 3y - 3 )2 - ( 4x + 6y + 2 )2
= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]
= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )
= ( -x - 3y - 5 )( 7x + 9y - 1 )
l) -4x2 + 12xy - 9y2 + 25
= 25 - ( 4x2 - 12xy + 9y2 )
= 52 - ( 2x - 3y )2
= ( 5 - 2x + 3y )( 5 + 2x - 3y )
m) x2 - 2xy + y2 - 4m2 + 4mn - n2
= ( x2 - 2xy + y2 ) - ( 4m2 - 4mn + n2 )
= ( x - y )2 - ( 2m - n )2
= ( x - y - 2m + n )( x - y + 2m - n )
Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2
a) \(\left(2x+3\right)^2-3\left(x-4\right)\left(x+4\right)=\left(x-2\right)^2+1\)
\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)
\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)
\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)
\(\Leftrightarrow16x+52=0\)
\(\Leftrightarrow x=-\frac{13}{4}\)
b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\)Xem lại đề !
c) \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-x^2-x+12=5x\)
\(\Leftrightarrow-2x+12=5x\)
\(\Leftrightarrow7x-12=0\)
\(\Leftrightarrow x=\frac{12}{7}\)
d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)
\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)
\(\Leftrightarrow28x+3x-1=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow x=\frac{1}{31}\)
`a, 3x(1 + 4y) - 5y(1 - 3x)`
`= 3x + 12xy - 5y + 15xy`
`= 3x + (12xy + 15xy) - 5y`
`= 3x + 27xy - 5y`
`b, -3xy(4 -2xy+4x)`
`= -3xy . 4 + 3xy . 2xy - 3xy . 4x`
`= -12xy + 6x^2y^2 - 12x^2y`
`c, (4-2x)(12-4x-6y^2)`
`= 48-16x - 24y^2 - 24x + 8x^2 + 12xy^2`
`= 48+ (-16x - 24x) - 24y^2 + 8x^2 + 12xy^2`
`= 48 - 40x + 24y^2 + 8x^2 + 12xy^2`
`d, (5 + 2y)(-4x - 6xy - 7x)`
`= 5 . (-4x) - 5 . 6xy - 5 . 7x - 2y . 4x - 2y . 6xy - 2y . 7x`
`= -20x - 30xy - 35x - 8xy - 12xy^2 -14x`
`= (-20x - 35x - 14x) + (-30xy - 8xy) - 12xy^2`
`= -69x - 38xy - 12xy^2`
`e, (4xy^3 - 6xy^2 - 12xy^4)/(-2xy)`
`= 4xy^3 : (-2xy) + 6xy^2 : 2xy + 12xy^4 : 2xy`
`= -2y^2 + 3y + 6y^3`