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a) x3 - 1 + 5x2 - 5 + 3x - 3
= (x - 1)(x2 + x + 1) + 5(x - 1)(x + 1) + 3(x - 1)
= (x - 1)(x2 + x + 1 + 5x + 5 + 3)
= (x - 1)(x2 + 6x + 9)
= (x - 1)(x + 3)2
b) (x + 1)(x + 2)(x + 3)(x + 4) + 1
= (x2 + 4x + x + 4)(x2 + 3x + 2x + 6) + 1
= (x2 + 5x + 4)(x2 + 5x + 6) + 1 (1)
Đặt t = x2 + 5x + 5
=> x2 + 5x + 4 = t - 1
x2 + 5x + 6 = t + 1
(1) = (t - 1)(t + 1) + 1
= t2 - 1 + 1
= t2 = (x2 + 5x + 5)2
c) x8 + x4 + 1
= x8 + x7 + x6 - x7 - x6 - x5 + x5 + x4 + x3 - x3 - x2 - x + x2 + x + 1
= x6(x2 + x + 1) - x5(x2 + x + 1) + x3(x2 + x + 1) - x(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x6 - x5 + x3 - x + 1)
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
\(2.\)
\(a.\)
\(x^2-25=0\)
\(\Rightarrow x^2-5^2=0\)
\(\Rightarrow\left(x-5\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(b.\)
\(5x^2-10x=0\)
\(\Rightarrow5x\left(x-10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=0\\x-10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
a, \(x^3-x^2=4x^2-8x+4\)
\(\Rightarrow x^3-x^2-4x^2+8x-4=0\)
\(\Rightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-2x-2x+4\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
b, \(\left(x-1\right)\left(x^2+x+1\right)=7\)
\(\Rightarrow x^3-1=7\Rightarrow x^3=2^3\Rightarrow x=2\)
c, \(2\left(x+5\right)-x^2-5x=0\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d, \(x^2-3x=-2\)
\(\Rightarrow x^2-x-2x+2=0\)
\(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) \(\left(x-9\right)\left(x-7\right)+1\)
\(=x^2-16x+63+1\)
\(=x^2-16x+64\)
\(=\left(x-8\right)^2\)
b) \(x^3+2x^2-3x-6\)
\(=x^2\left(x+2\right)-3x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-3x\right)\)
\(=x\left(x+2\right)\left(x-3\right)\)
c) \(x^2-y^2+xz-yz\)
\(=x\left(x+z\right)-y\left(y+z\right)\)
\(=\left(x-y\right)\left(y+z\right)\)
d) \(x^3-x+3x^2y+y^3-y\)
botay:(
a) \(x^3-1+5x^2-5+3x-3\)
= \(x^3+5x^2+3x-9\)
= \(x^3-x^2+6x^2-6x+9x-9\)
= \(x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+6x+9\right)\)
= \(\left(x-1\right)\left(x-3\right)^2\)
b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
= \(\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
= \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\) (1)
Đặt \(x^2+5x+4=a\)
Đa thức (1) \(\Leftrightarrow a\left(a+2\right)+1\)
= \(a^2+2a+1=\left(a+1\right)^2=\left(x^2+5x+4+1\right)^2\)
= \(\left(x^2+5x+6\right)^2\)
c) \(x^8+x^4+1\)
Ta thấy \(\left\{{}\begin{matrix}x^8\ge0\\x^4\ge0\\1>0\end{matrix}\right.\) \(\Rightarrow x^8+x^4+1\ge1\)
\(\Rightarrow\) Không phân tích thành nhân tử đc.
d) \(x^3+x^2+4\)
= \(x^3+2x^2-x^2+4\)
= \(x^2\left(x-2\right)-\left(x^2-4\right)\)
= \(x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\)
= \(\left(x-2\right)\left(x^2-x-2\right)\)