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Bài2: phân tích đa thức thành nhân tử
\(a,x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(y+x-2\right)\)
\(b,x^3-5x^2+x-5\)
\(=x^2\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+x-5\right)\left(x-x-5\right)\)
\(c,x^2-2xy+y^2-9\)
\(=\left(x^2-y^2\right)-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
chúc bạn học tốt !
a) A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
A = 6x^2 + 33x - 10x - 55 - 6x^2 - 23x - 21
A = -76
b) B = 4x(3x - 2) - 3x(4x + 1)
B = 12x^2 - 8x - 12x^2 - 3x
B = -11x
c) C = (x + 3)(x - 2) - (x - 1)^2
C = x^2 + x - 6 - x^2 + 2x - 1
C = 3x - 7
Bài 1: Thực hiện phép tính
a) 3x(2x2 - 5x + 9) = \(6x^3-15x^2+27x\)
b) 5x(x2-xy+1) = \(5x^3-5xy+5x\)
c) -2/3x2y(3xy-x2+y) = \(-2x^3y^2+\dfrac{2}{3}x^4y-\dfrac{2}{3}x^2y^2\)
2) Thực hiện phép tính
a) (5x-2y) (x2-xy+1) = \(5x^3+5x-7y-2x^3y+2xy^2\)
b) (x+3y)(x2-2xy+y) = \(x^3-x^2y+xy+6xy^2+y^2\)
c) (3x-5y) (4x+ 7y) = \(12x^2-xy-35y^2\)
Bài 3: Rút gọn các biểu thức sau(bằng cách khai triển hằng đẳng thức):
a) (x+y)2+(x-y)2
= \(x^2+2xy+y^2+x^2-2xy+y^2\)
= \(\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
= \(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) (x+2)(x-2)-(x-3)(x+1)
= \(x^2-4\) - \(\left(x^2-2x-3\right)\)= \(x^2-4-x^2+2x+3\)
= \(\left(x^2-x^2\right)+2x+\left(-4+3\right)\)=\(2x-1\)
c) (x-2)(x+2)-(x-2)2
=>\(x^2-4-\left(x^2-2.x.2+2^2\right)=x^2-4-x^2-4x+4=\left(x^2-x^2\right)+\left(-4+4\right)-4x=-4x\)
d) (2x+y)(4x2-2xy+y2)-(2x-y)(4x2+2xy+y2)
= \(8x^3+y^3-\left(8x^3-y^3\right)\)
= \(8x^3+y^3-8x^3+y^3\)
= \(\left(8x^3-8x^3\right)+\left(y^3+y^3\right)\)= \(2y^3\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
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
Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
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)
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)