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Bài 2:
1) \(7x^2+2x=0\)
\(\Leftrightarrow x\left(7x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\7x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{7}\end{matrix}\right.\)
2) \(2x\left(x-9\right)+5\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{5}{2}\end{matrix}\right.\)
3) \(x^2+8x+16=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Bài 1:
2) \(24x-18y+30=6\left(4x-3y+5\right)\)
5) \(x^2+14x+49=\left(x+7\right)^2\)
6) \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)
i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
a) 2x(y-2009)+5y(y-2009)
=(y-2009)(2x+5y)
b) 35x(y-8)-14y(8-y)
=35x(y-8)+14y(y-8)
=7(y-8)(5x+2y)
c) 15x^2 +10xy =5x(3x+2y)
d)24x-18y+30= 6(4x+3y+10)
e) x^2 +14x+49
= x^2 +2.7.x+ 7^2
=(x+7)^2
i) x^2 -x - y^2 - y
=(x^2 -y^2)-(x+y)
=(x-y)(x+y)-(x+y)
=(x+y)(x-y-1)
k) x^2 -2xy+y^2 -z^2
=(x^2 -2xy+y^2) -z^2
=(x-y)^2 -z^2
= (x-y-z)(x-y+z)
a) \(2x\left(y-2009\right)+5y\left(y-2009\right)\) \(=\left(y-2009\right)\left(2x+5y\right)\)
b) \(35x(y-8)-14y(8-y)\) \(=35x\left(y-8\right)+14y\left(y-8\right)\)
\(=\left(y-8\right)\left(35x+14y\right)\)
\(=\left(y-8\right).7\left(5x+2y\right)\)
c) \(15x^2+10xy=5x\left(3x+2y\right)\)
d) \(24x-18y+30=3\left(8x-6y+10\right)\)
e) \(x^2+14x+49=x^2+2.7.x+7^2\)
\(=\) \((x+7)^2\)
g) \(27x^3+y^3\) \(=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
h) \(8x^3-\dfrac{1}{125}y^3=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)
i) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
k) \(x^2-2xy+y^2-z^2=\left(x^2-2xy+y^2\right)-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
- HỌC TỐT NHA BẠN YÊU ^^ -
g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)
f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)\)
\(=6x^2\left(x+1\right)\)
f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=6x^2\left(x+1\right)\)
g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)
câu f có ( x+1) là nhân tử chung
câu g đổi dấu - thành + thì (y-x) sẽ thành (x-y)
e) \(8\left(x+3y\right)-16x\left(x+3y\right)=\left(x+3y\right)\left(8-16x\right)=8\left(x+3y\right)\left(1-2x\right)\)
f) \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=\left(x+1\right)\left(4x^2+2x^2\right)=6x^2\left(x+1\right)\)
g) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)
\(2x^2y^3-\frac{x}{4}-4y^6\)
đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được
\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)
\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)
cái này có những biểu thức ko dùng được hđt hoặc đặt nhân tử chung =))) anh nghĩ đề em nên thay chữ và thành hoặc
1, \(15x^2+10xy=5x\left(3x+2y\right)\)
2, \(24x-18y+30=6\left(4x-3y+5\right)\)
3, \(2x\left(y-2009\right)+5y\left(y-2009\right)=\left(y-2009\right)\left(2x+5y\right)\)
4, \(35x\left(y-8\right)-14y\left(8-y\right)=\left(y-8\right)\left(35x+14y\right)=7\left(5x+2y\right)\left(y-8\right)\)
5, \(x^2+14x+49=x^2+2.7x+7^2=\left(x+7\right)^2\)
6, \(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)
Trả lời:
1, 15x2 + 10xy = 5x ( 3x + 2y )
2, 24x - 18y + 30 = 6 ( 4x - 3y + 5 )
3, 2x ( y - 2009 ) + 5y ( y - 2009 ) = ( y - 2009 )( 2x + 5y )
4, 35x ( y - 8 ) - 14y ( 8 - y ) = 35x ( y - 8 ) + 14y ( y - 8 ) = ( y - 8 )( 35x + 14y ) = 7 ( y - 8 )( 5x + 2y )
5, x2 + 14x + 49 = ( x + 7 )2
6, 9x2 - 4 = ( 3x - 2 )( 3x + 2 )