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a) x2- 2x - 4y2 - 4y = (x2 - 2x + 1) - (4y2 + 4y + 1) = (x - 1)2 - (2y + 1)2 = (x - 1 - 2y - 1)(x - 1 + 2y + 1) = (x - 2y - 2)(x + 2y)
b) x3 - 4x2 + 12x - 27 = (x3 - 3x2) - (x2 - 3x) + (9x - 27) = x2(x - 3) - x(x - 3) + 9(x - 3) = (x2 - x + 9)(x - 3)
d) x4 - 2x3 + 2x - 1 = (x4 - 2x3 + x2) - (x2 - 2x + 1) = (x2 - x)2 - (x - 1)2 = (x2 - x - x + 1)(x2 - x + x - 1)
= (x2 - 2x + 1)(x2 - 1) = (x - 1)2(x - 1)(x + 1) = (x - 1)3(x + 1)
e) x4 + 2x3 - 4x - 4 = (x4 + 2x4 + x2) - (x2 + 4x + 4) = (x2 + x)2 - (x + 2)2 = (x2 + x - x - 2)(x2 + x + x + 2) = (x2 - 2)(x2 + 2x + 2)
phân tích các đa thức sau thành nhân tử:
a) \(\left(ab-1\right)^2+\left(a+b\right)^2\)
Ta thấy \(\left\{{}\begin{matrix}\left(ab-1\right)^2\ge0\\\left(a+b\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(ab-1\right)^2+\left(a+b\right)^2>0\) nên k phân tích thành nhân tử đc.
b) \(x^3+2x^2+2x+1\)
= \(x^3+x^2+x^2+x+x+1\)
= \(x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27\)
= \(x^3-3x^2-x^2+3x+9x-27\)
= \(x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
= \(\left(x-3\right)\left(x^2-x+9\right)\)
d) \(x^4+2x^3+2x^2+2x+1\)
= \(x^4+x^3+x^3+x^2+x^2+x+x+1\)
= \(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
= \(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
= \(\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
= \(\left(x+1\right).\left(x+1\right)\left(x^2+1\right)\)
= \(\left(x+1\right)^2\left(x^2+1\right)\)
a, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
\(=a^2b^2-2ab+1+a^2+2ab+b^2\)
\(=a^2b^2+a^2+b^2+1=a^2.\left(b^2+1\right)+\left(b^2+1\right)\)
\(=\left(b^2+1\right).\left(a^2+1\right)\)
b, \(x^3+2x^2+2x+1\)
\(=x^3+x^2+x^2+x+x+1\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2.\left(x-3\right)-x.\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right).\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=x^4-x^3-x^3+x^2-x^2+x+x-1\)
\(=x^3.\left(x-1\right)-x^2.\left(x-1\right)-x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right).\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right).\left[x^2.\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right).\left(x-1\right).\left(x^2-1\right)=\left(x-1\right)^2\left(x^2-1\right)\)
e, \(x^4+2x^3+2x^2+2x+1\)
\(=x^4+x^3+x^3+x^2+x^2+x+x+1\)
\(=x^3.\left(x+1\right)+x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+x^2+x+1\right)\)
\(=\left(x+1\right).\left[x^2.\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right).\left(x+1\right).\left(x^2+1\right)=\left(x+1\right)^2\left(x^2+1\right)\)
Chúc bạn học tốt!!!
a) \(x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)=0\)
\(TH1:x+1=0\Leftrightarrow x=-1\)
\(TH2:x^2+x+1=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)
Mà \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)nên loại TH2
Vậy x = 1
Câu a), x = -1 nha, kết luận nhầm
b) \(x^3-4x^2+12x-27=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+9\right)-4x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-7x+9\right)=0\)
\(TH1:x-3=0\Leftrightarrow x=3\)
\(TH2:x^2-7x+9=0\)
\(\cdot\Delta=\left(-7\right)^2-4.9=13\)
Vậy pt của TH2 có 2 nghiệm phân biệt
\(x_1=\frac{7+\sqrt{13}}{2}\);\(x_2=\frac{7-\sqrt{13}}{2}\)
c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)
\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)
Tương tự với a ; b
1.A =( x-3)( x+3) + 15 - x2
A=X2-3X+3X+15-X3
A=15-X
2.B=(X -1) (X2+X+1) - X (X2+2) + 2X
B=X3+ X2+ X - X2 - X - 1 - X3 - 2X + 2X
B= -1
3.C=(2X - 1 ) (4X2 + 2X + 1) - X ( 8 X 2 + 1 ) + X
C=8X3 - 4X2 +4X2 - 2X +2 X - 1 - 8X22 - X + X
C=8X3 - 1 - 8X22
MK CHỈ LM ĐC TỚI ĐÓ THUI SAI CHỖ NÀO ĐỪNG TRÁCH VÌ MK YẾU PHẦN NÀY
Ta có : x3 - 4x2 + 12x - 27
= (x3 - 27) - (4x2 - 12x)
= (x3 - 33) - 4x(x - 3)
= (x - 3)(x2 + 3x + 9) - 4x(x - 3)
= (x - 3) (x2 + 3x + 9 - 4x)
= (x - 3)(x2 - x + 9)
a) \(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)
\(x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
c) \(\left(x^3-27\right)-4x\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)