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\(2.\)
\(a.\)
\(x^2-25=0\)
\(\Rightarrow x^2-5^2=0\)
\(\Rightarrow\left(x-5\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(b.\)
\(5x^2-10x=0\)
\(\Rightarrow5x\left(x-10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=0\\x-10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Bài 1:
a) \(\left(x+3\right)^2+x\left(x-2\right)=2x^2\)
\(x^2+6x+9+x^2-2x-2x^2=0\)
\(4x+9=0\)
\(x=\frac{-9}{4}\)
b) \(5x\left(x-4\right)-x+4=0\)
\(5x\left(x-4\right)-\left(x-4\right)=0\)
\(\left(x-4\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{5}\end{cases}}}\)
Bài 2:
a) \(x^2-4x=x\left(x-4\right)\)
b) \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)
c) \(x^2-y^2+2y-1\)
\(=x^2-\left(y^2-2y+1\right)\)
\(=x^2-\left(y-1\right)^2\)
\(=\left(x-y+1\right)\left(x+y-1\right)\)
d) \(x^2-11x+18\)
\(=x^2-2x-9x+18\)
\(=x\left(x-2\right)-9\left(x-2\right)\)
\(=\left(x-2\right)\left(x-9\right)\)
(x + 3)2 + x(x - 2) = 2x2
x2 + 6x + 9 + x2 - 2x = 2x2
<=> 2x2 + 4x + 9 = 2x2
<=> 4x = -9
<=> x = -9/4
Bài 1:
a) \(x^2-y^2+10x+25\)
\(=\left(x^2+10x+25\right)-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+y+5\right)\left(x-y+5\right)\)
b) \(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^4+4y^4\)
\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
e) \(7x^2-10xy+3y^2\)
\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)
\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)
f) Sửa đề \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)
\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)
\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
a) -25x6 - y8 + 10x3y4 = -25x6 + 10x3y4 - y8
= - ( 25x6 - 10x3y4 + y8 )
= - [ ( 5x3 )2 - 2 . 5x3y4 + ( y4 )2 ]
= - ( 5x3 - y4 )2
b) \(\dfrac{1}{4}\)x2 - 5xy + 25y2 = (\(\dfrac{1}{2}\)x)2 - 2 . \(\dfrac{1}{2}\) x . 5y + ( 5y )2
= (\(\dfrac{1}{2}\) x - 5y )2
c) ( x - 5 )2 - 16 = ( x - 5 )2 - 42
= ( x - 5 - 4 ) . ( x - 5 + 4 )
= ( x - 9 ) . ( x - 1 )
d) 25 - ( 3 - x )2 = 52 - ( 3 - x )2
= ( 5 - 3 + x ) . ( 5 + 3 - x )
= ( x + 2 ) . ( 8 - x )
a, = (x^2+10x+25)-y62 = (x+5)^2-y^2 = (x+5-y).(x+5+y)
b, = xy.(x-y)
c, = (x-y).(x+y)+5.(x-y) = (x-y).(x+y+5)
k mk nha
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(1,3x-24y=3\left(x-8y\right)\)
\(2,6x^3y^2-12x^2y^2-3x^2y=3x^2y\left(2xy-4y-1\right)\)
\(3,7x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(7x-8\right)\)
...(tương tự)
\(10,5x-5y+x^2-xy=5\left(x-y\right)+x\left(x-y\right)=\left(x-y\right)\left(x+5\right)\)
\(11,x^2+2xy+y^2-16=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
2D
6
\(x^3+125=\left(x+5\right)\left(x^2-5x+25\right)\)
A là đa thức bậc 1
=>A=x+5
=>B=x^2-5x+25
=>Chọn A
Câu 2. M có bậc 2 + 7 = 9
Chọn D
Câu 6. x³ + 125 = x³ + 5³ = (x + 5)(x² - 5x + 25)
Chọn A