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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,\(=x^4-3x^3+3x^3-9x^2-4x^2+12x-12x+36\)
\(=x^3\left(x-3\right)+3x^2\left(x-3\right)-4x\left(x-3\right)-12\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+3x^2-4x-12\right)\)
\(=\left(x-3\right)[x^2\left(x+3\right)-4\left(x+3\right)]\)
\(=\left(x^2-9\right)\left(x^2-4\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)
\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)
\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)
\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)
\(=\left(x-3\right)\left(2x^2-11x\right)\)
\(=x\left(x-3\right)\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)
\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
Bài 1 :
a) \(x^4-4x^2-4x-1\)
\(=x^4-\left(4x^2+4x+1\right)\)
\(=x^4-\left(2x+1\right)^2\)
\(=\left(x^2-2x-1\right)\left(x^2+2x+1\right)\)
b) \(x^2+2x-15\)
\(=x^2+2x+1-16\)
\(=\left(x+1\right)^2-4^2\)
\(=\left(x+1+4\right)\left(x+1-4\right)=\left(x+5\right)\left(x-3\right)\)
c) \(x^3y-2x^2y^2+5xy\)
\(=xy\left(x^2-2xy+5\right)\)
B2:
a) \(2\left(x-1\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=2\left(x^2-2x+1\right)-\left(4x^2-9\right)\)
\(=2x^2-4x+2-4x^2+9\)
\(=-2x^2-4x+11\)
b) \(\left(x+3\right)^2-2\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(x+3-x+3\right)^2=6^2=36\)
c) \(4\left(x-1\right)\left(x+3\right)+5\left(2x+1\right)^2-2\left(5-3x\right)^2\)
\(=4\left(x^2+2x-3\right)+5\left(4x^2+4x+1\right)-2\left(9x^2-30x+25\right)\)
\(=4x^2+8x-12+20x^2+20x+5-18x^2+60x-50\)
\(=6x^2+88x-57\)
Bài 2.
a) x(x-2)-x+2=0
<=> x2-2x-x+2=0
<=> x2-3x+2=0
<=> x2-x-2x-2=0
<=> x(x-1)-2(x-1)=0
<=> (x-1)(x-2)=0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
b) x2(x2+1)-x2-1=0
<=> x4+x2-x2-1=0
<=> x4-1=0
<=> x4=1
<=> x=\(\pm\)1
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