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1. x2 - 6x + 9=(x-3)2
2. 25 + 10x + x2=(x+5)2
3. \(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5.x3 + 8y3=(x+8y)(x2-8xy+64y2)
6.8y3 -125=(2y-5)(4y2+10y+25)
7.a6-b3=(a2-b)(a4+a2b+b2)
8 x2 - 10x + 25=(x-2)2
1) \(x^2-6x+9=\left(x-3\right)^2\)
2) \(25+10x+x^2=\left(5+x\right)^2\)
3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)
7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
8) \(x^2-10x+25=\left(x-5\right)^2\)
9) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
a: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
b: \(27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
c: \(y^6+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
d: \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
x 3 – 6 x 2 y + 12 x y 2 – 8 y 3 = x 3 – 3 . x 2 . ( 2 y ) + 3 . x . ( 2 y ) 2 – ( 2 y ) 3 = ( x – 2 y ) 3
đáp án cần chọn là: D
Câu 17:
Xét ΔADC có OE//DC
nên \(\dfrac{OE}{DC}=\dfrac{AO}{AC}\left(1\right)\)
Xét ΔBDC có OH//DC
nên \(\dfrac{OH}{DC}=\dfrac{BO}{BD}\left(2\right)\)
Xét ΔOAB và ΔOCD có
\(\widehat{OAB}=\widehat{OCD}\)(hai góc so le trong, AB//CD)
\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)
Do đó: ΔOAB đồng dạng với ΔOCD
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)
=>\(\dfrac{OC}{OA}=\dfrac{OD}{OB}\)
=>\(\dfrac{OC}{OA}+1=\dfrac{OD}{OB}+1\)
=>\(\dfrac{OC+OA}{OA}=\dfrac{OD+OB}{OB}\)
=>\(\dfrac{AC}{OA}=\dfrac{BD}{OB}\)
=>\(\dfrac{OA}{AC}=\dfrac{OB}{BD}\left(3\right)\)
Từ (1),(2),(3) suy ra \(\dfrac{OE}{DC}=\dfrac{OH}{DC}\)
=>OE=OH
Câu 15:
a: \(3x\left(x-1\right)+x-1=0\)
=>\(3x\left(x-1\right)+\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b: \(x^2-6x=0\)
=>\(x\cdot x-x\cdot6=0\)
=>x(x-6)=0
=>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
a ) \(x^3+8y^3=x^3+\left(2y\right)^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
b ) \(a^6-b^3=\left(a^2\right)^3-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c ) \(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
d ) \(8z^3+27=\left(2z\right)^3+3^3=\left(2z+3\right)\left(4z^2-6z+9\right)\)
a) x3 + 8y3 = x3 + (2y)3 = (x+2y)(x2+2xy+4y2)
b) a6 - b3 = (a2)3 - b3 = (a2-b)(a4 + a2b + b2)
c) 8y3 - 125 = (2y)3 - 53 = (2y - 5)(4y2 + 10y + 25)
d) 8x3 + 27 = (2z)3 + 33 = (2z + 3)(4z2 - 6x + 9)