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a)
\(x^3-5x^2+6x\\ \Leftrightarrow x\cdot\left(x^2-5x+6\right)\\ \Leftrightarrow x\cdot\left(x^2-2x-3x+6\right)\\ \Leftrightarrow x\cdot\left[x\cdot\left(x-2\right)-3\cdot\left(x-2\right)\right]\\ \Leftrightarrow x\cdot\left(x-3\right)\cdot\left(x-2\right)\)
b)
\(x^2-3xy+2y^2\\ \Leftrightarrow x^2-xy-2xy+2y^2\\ \Leftrightarrow x\cdot\left(x-y\right)-2y\cdot\left(x-y\right)\\ \Leftrightarrow\left(x-2y\right)\cdot\left(x-y\right)\)
c)
\(-4x^2+10x-4\\ \Leftrightarrow-2\cdot\left(2x^2-5x+2\right)\\ \Leftrightarrow-2\cdot\left(2x^2-x-4x+2\right)\\ \Leftrightarrow-2\cdot\left[x\cdot\left(2x-1\right)-2\cdot\left(2x-1\right)\right]\\ \Leftrightarrow-2\cdot\left(x-2\right)\cdot\left(2x-1\right)\)
d)
\(x^3+2x^2y-xy^2-2y^3\\ \Leftrightarrow x^2\cdot\left(x+2y\right)-y^2\cdot\left(x+2y\right)\\ \Leftrightarrow\left(x+2y\right)\cdot\left(x^2-y^2\right)\\ \Leftrightarrow\left(x+2y\right)\cdot\left(x+y\right)\cdot\left(x-y\right)\)
cái này là phép toán dễ mà, chỉ cần nắm vũng kiến thức trong chương 1 sách lớp 8 là đc có j đâu?
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
a) \(4x^3y-12x^2y^3-8x^4y^3\)
\(=4x^2y\left(x-3y^2-2x^2y^2\right)\)
b) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
c) \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-y-1\right)\left(x+y-1\right)\)
d) \(x\left(x-2y\right)+3\left(2y-x\right)\)
\(=x\left(x-2y\right)-3\left(x-2y\right)\)
\(=\left(x-3\right)\left(x-2y\right)\)
e) \(x^2+4\)
\(=\left(x^4+4x^2+4\right)-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
f) \(5x^2-7x-6\)
\(=\left(5x^2-10x\right)+\left(3x-6\right)\)
\(=5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(5x+3\right)\left(x-2\right)\)
a) ( -5x2 +3xy + 7) + ( -6x2y + 4xy2 - 5)=4*x*y^2-6*x^2*y+3*a*x*y-5*a*x^2+7*a-5
b) ( 2,4x3 - 10x2y) + (7x2y - 2,4x3 + 3xy2)=3*x*y^2-3*x^2*y
c) ( 15x2y - 7xy2 - 6y2) + (2x2 - 12x2y + 7xy2)=-6*y^2+3*x^2*y+2*x^2
d) ( 4x2 + x2y - 5y3) + (5/3 x3 - 6xy2 - x2y) + (x3/3 + 10y3) + ( 6y3-15xy2 - 4x2y - 10x3)=11*y^3-21*x*y^2-4*x^2*y-8*x^3+4*x^2
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