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a) (2x+y)3
c)(x2-y2)(x4+x2y2+y4)
d)-x3+9x2-27x+27
<=> -(x3-9x2+27x-27)
<=>-(x-3)3
a. \(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
\(=2x^3+6xy^2\)
\(=2x\left(x^2+6y^2\right)\)
b. \(x^3-y^3+2x^2-2y^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)
c. \(x^3-y^3-3x^2+3x-1\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2+y^2+xy-2x-y+1\right)\)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
\(3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2\left(x-2y+3\right)\)
\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
\(12x^2y-18xy^2-3xy^2=3xy\left(4x-6y-y\right)\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(y\left(x-z\right)+7\left(z-x\right)=y\left(x-z\right)-7\left(x-z\right)=\left(x-z\right)\left(y-7\right)\)
\(27x^2\left(y-1\right)-9x^3\left(1-y\right)=27x^2\left(y-1\right)+9x^3\left(y-1\right)=9x^2\left(y-1\right)\left(3-x\right)\)
a. \(1-2y+y^2=\left(1-y\right)^2\)
b. \(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
c. \(1-4x^2=\left(1+2x\right)\left(1-2x\right)\)
d. \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
e. \(27+27x+9x^2+x^3=\left(x+3\right)^3\)
f, \(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
g, \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(\left(a\right)1-2y+y^2\)
\(\Leftrightarrow y^2-2y+1\)
\(\Leftrightarrow\left(y-1\right)^2\)
\(\left(b\right)\left(x+1\right)^2-25\)
\(\Leftrightarrow\left(x+1\right)^2-5^2\)
\(\Leftrightarrow\left(x-4\right)\left(x+6\right)\)
\(\left(c\right)1-4x^2\)
\(\Leftrightarrow1-\left(2x\right)^2\)
\(\Leftrightarrow\left(1-2x\right)\left(1+2x\right)\)
\(\left(d\right)8-27x^3\)
\(\Leftrightarrow2^3-\left(3x\right)^3\)
\(\Leftrightarrow\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(\left(e\right)27+27x+9x^2+x^3\)
\(\Leftrightarrow\left(x+3\right)^3\)
\(\left(f\right)8x^3-12x^2y+6xy^2-y^3\)
\(\Leftrightarrow\left(2x\right)^3-12x^2y+6xy^2-y^3\)
\(\Leftrightarrow\left(2x-y\right)^3\)
\(\left(g\right)x^3+8y^3\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
a) 1 - 2y + y2
= (1-y)2
b) ( x + 1 )2 - 25
=( x + 1 )2 - 52
=(x+1+5)(x+1-5)
Bài 1 :
a) (3a+4b)3+(3a-4b)3-48a2b2
=27a3+108a2b+144ab2+64b3+27a3-108a2b+144ab2-64b3-48a2b2
=54a3+288ab2-48a2b2
=2a(27a2+144b2-24ab)
b) (5x+2y)(5x-2y)+(2x-y)3+(2x+y)3
=25x2-4y2+8x3-12x2y+6xy2-y3+8x3+12x2y+6xy2+y3
=16x3+25x2-y2+12xy2
=x2(16x+25)-y2(1-12x)
Bài 2 :
\(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
b)\(x^3-4x^2+3x=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)
c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn
d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)
\(\Leftrightarrow27x^3-54x^2-27x=0\)
\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
#H
a/ 27x^3 + 54x^2y+36xy^2+8y^3=(3x+27)^3
b/ 8x^3+12x^2y+6xy^2+y^3=(2x+y)^3
c/ 3x^3-12x^2y+6xy^2-y^3=(2x-y)^3
d/ x^3-6x^2y+12xy^2-8y^3=(x-2y)^3