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Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
9(a + b)2 - (a + b) = (a + b)[9(a + b) - 1]
(mx + my) + (3x + 3y) = m(x + y) + 3(x + y) = (m + 3)(x + y)
(12xy) - 6x - (2y - 1) = 6x(2y - 1) - (2y - 1) = (6x - 1)(2y - 1)
(7xy2 - 5x2y) + (5x - 7y) = xy(7y - 5x) + (5x - 7y) = -xy(5x - 7y) + (5x - 7y) = (-xy + 1)(5x - 7y)
2x(x - y) - (4x - 4y) = 2x(x - y) - 4(x - y) = (2x - 4)(x - y)
a) 9( a + b )2 - ( a + b ) = ( a + b )[ 9( a + b ) - 1 ]
b) ( mx + my ) + ( 3x + 3y ) = m( x + y ) + 3( x + y ) = ( m + 3 )( x + y )
c) 12xy - 6x - ( 2y - 1 ) = 6x( 2y - 1 ) - ( 2y - 1 ) = ( 6x - 1 )( 2y - 1 )
d) ( 7xy2 - 5x2y ) + ( 5x - 7y ) = xy( 7y - 5x ) + ( 5x - 7y ) = -xy( 5x - 7y ) + ( 5x - 7y ) = ( -xy + 1 )( 5x - 7y )
e) 2x( x - y ) - ( 4x - 4y ) = 2x( x - y ) - 4( x - y ) = ( 2x - 4 )( x - y )
a/Dùng hằng đẳng thức A2-B2=(A+B)(A-B) phân tích được ngay
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
=\(\left(3x-2y+3\right)\left(4-x-4y\right)\)
b/Chắc chỉ phân tích hằng đẳng thức (A-B)2=A2-2ab+B2
\(49\left(y-4\right)^2-9y^2-3y-36=49y^2-392y+784-9y^2-3y-36\)
\(=40y^2-395y+748\)
Mình dùng biệt thức cho ra nghiệm vô tỉ, không biết cho phải tại mình tính sai hay đề thiếu nữa
c/Khai triển biểu thức ban đầu ta được
\(x\left(x-y\right)+y\left(y-x\right)=x^2-xy+y^2-xy=x^2-2xy+y^2=\left(x-y\right)^2\)
Mình nghĩ bạn ghi đề sai, đề đúng theo mình là:
\(x^2y^2\left(x-y\right)+y^2z^2\left(y-z\right)+z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(x-y\right)-y^2z^2\text{[}\left(x-y\right)+\left(z-x\right)\text{]}+z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(x-y\right)-y^2z^2\left(x-y\right)-y^2z^2\left(z-x\right)+z^2x^2\left(z-x\right)\)
\(=\left(x-y\right)\left(x^2y^2-y^2z^2\right)+\left(z-x\right)\left(z^2x^2-y^2z^2\right)\)
\(=\left(x-y\right).y^2\left(x+z\right)\left(x-z\right)+\left(z-x\right).z^2\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x-z\text{ }\right)\text{[}y^2.\left(x+z\right)-z^2\left(x+y\right)\text{]}\)
\(=\left(x-y\right)\left(z-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-y\right)\left(z-x\right)\text{[}\left(y^2x-z^2x\right)+\left(y^2z-z^2y\right)\text{]}\)
\(=\left(x-y\right)\left(z-x\right)\text{[}x.\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\text{]}\)
\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(xy+x\text{z}+yz\right)\)
\(\left(2x-y\right)\left(x-y\right)-\left(3y-4x\right)^2+\left(y-2x\right)\left(2y-3x\right)\)
=(2x-y)(x-y)-(2x-y)(2y-3x)-(4x-3y)2
=(2x-3y)(x-y-2y+3x)-(4x-3y)2
=(2x-3y)(4x-3y)-(4x-3y)2
=(4x-3y)(2x-3y-4x+3y)
=(4x-3y))(-2x)