Phân tích đa thức thành nhân tử
- x3+2x-3
- x3+5x2+8x+4
- x3-7x+6
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a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
\(=x^3-x+7x+7=x\left(x-1\right)\left(x+1\right)+7\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x+7\right)\)
\(\text{x^3 – 5x^2 + 8x – 4 }\)
\(\text{= x^3 – 4x^2 + 4x – x^2 + 4x – 4}\)
\(\text{= x( x^2 – 4x + 4 ) – ( x^2 – 4x + 4 )}\)
\(\text{= ( x – 1 ) ( x – 2 )^2}\)
\(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x-4\\ =x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\\ =\left(x^2-4x+4\right)\left(x-1\right)\\ =\left(x-2\right)^2\left(x-1\right)\)
\(9a^3-13a+6=\left(9a^3-6a^2\right)+\left(6a^2-4a\right)-\left(9a-6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
\(x^4-4x^3+8x+3=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(5x^2+5x\right)+\left(3x+3\right)=x^3\left(x+1\right)-5x^2\left(x+1\right)+5x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^3-5x^2+5x+3\right)=\left(x+1\right)\left[\left(x^3-3x^2\right)-\left(2x^2-6x\right)-\left(x-3\right)\right]=\left(x+1\right)\left(x-3\right)\left(x^2-2x-1\right)\)
Sửa đề: x^3+6x^2+11x+6
=x^3+x^2+5x^2+5x+6x+6
=(x+1)(x^2+5x+6)
=(x+1)(x+2)(x+3)
\(A=x^2-y^2+7x+7y\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+7\right)\)
\(B=4x^3-4x^2+x\)
\(=x\left(4x^2-4x+1\right)\)
\(=x\left(2x-1\right)^2\)
\(C=x^2-6xy+9y^2-9\)
\(=\left(x-3y\right)^2-9\)
\(=\left(x-3y-3\right)\left(x-3y+3\right)\)
A=\(x^2+7x+7y-y^2=\left(x^2-y^2\right)+\left(7x+7y\right)=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)=\left(x+y\right)\left(x-y+7\right)\)
B=\(4x^3-4x^2+x=x\left(4x^2-4x+1\right)=x\left(2x-1\right)^2\)
C=\(x^2+9y^2-9-6xy=\left(x^2-6xy+9y^2\right)-9=\left(x-3y\right)^2-3^2=\left(x-3y-3\right)\left(x-3y+3\right)\)
\(x^3+2x-3\)
\(=x^3-x+3x-3\)
\(=x\left(x^2-1\right)+3\left(x-1\right)\)
\(=x\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x\left(x+1\right)+3\right)\)
\(x^3+5x^2+8x+4\)
\(=\left(x+1\right)\left(x+2\right)\)
Cây cuối tương tự câu đầu thôi:
\(x^3-7x+6\)
\(=x^3-x-6x+6\)
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