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a) \(2x^2+5x-18\)
\(=2x^2-4x+9x-18\)
\(=2x\left(x-2\right)+9\left(x-2\right)\)
\(=\left(x-2\right)\left(2x+9\right)\)
b) \(4x^2-17x+15\)
\(=4x^2-12x-5x+15\)
\(=4x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(4x-5\right)\)
c) \(-8x^2+10x+7\)
\(=-8x^2-4x+14x+7\)
\(=-4x\left(2x+1\right)+7\left(2x+1\right)\)
\(=\left(2x+1\right)\left(-4x+7\right)\)
d) \(7x^2-30x+8\)
\(=7x^2-28x-2x+8\)
\(=7x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(7x-2\right)\)
e) \(-x^3+11x^2-30x\)
\(=x\left(-x^2+11x-30\right)\)
\(=x\left(-x^2+5x+6x-30\right)\)
\(=x\left[-x\left(x-5\right)+6\left(x-5\right)\right]\)
\(=x\left(x-5\right)\left(-x+6\right)\)
a) 2x\(^2\) + 5x - 18 = 2x\(^2\) + 9x - 4x - 18 = x(2x + 9) - 2(2x + 9) = (x-2)(2x-9)
b) 4x\(^2\) - 17x - 15 = 4x\(^2\) + 20x - 3x - 15 = 4x(x + 5 ) - 3(x + 5) = (4x - 3 )(x + 5)
c) -8x\(^2\) + 10x + 7 = -8x\(^2\) + 14x - 4x + 7 =-2x(4x - 7) - (4x - 7) = (-2x - 1)(4x - 7)
d) 7x\(^2\) - 30x + 8 = 7x\(^2\) + 2x + 28x + 8 = x(7x + 2) + 4(7x + 2) = (x + 4)(7x + 2)
e) - x\(^3\) + 11x\(^2\) - 30x = -x(x\(^2\) - 11x + 30) = -x(x\(^2\) - 5x - 6x + 30) = -x\(\left[x\left(x-5\right)-6\left(x-5\right)\right]\) = -x(x-6)(x-5)
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3
a) (7x + 4)2 - (7x + 4)(7x - 4)
= 49x2 + 56x + 16 - 49x2 + 16
= 56x + 32
b) (x - 2y)3 - 6xy(x - 2y)
= x3 - 6x2y + 12xy2 - 8y3 - 6x2y + 12xy2
= x3 - 12x2y + 24xy2 - 8y3
c) (3x + y)(9x2 - 3xy + y2) - (3xy)3 - 27x2y
= 27x3 + y3 - (3xy)3 - 27x2y
d) 5(x + 3)(x - 3) + (2x + 3)2 + (x - 6)2
= 5x2 - 45 + 4x2 + 12x + 9 + x2 - 12x + 36
= 10x2
e) (2x + 3)2 + (2x - 3)2 - 2(4x2 - 9)
= (2x + 3)2 + (2x - 3)2 - 2(2x - 3)(2x + 3)
= (2x + 3 - 2x + 3)2
= 62 = 36
g) (x + 2)3 + (x - 2)3 + x3 - 3x(x - 2)(x + 2)
= (x+2+x-2)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4) + x3 - 3x3 + 12x
= 2x(x2 + 8) + x3 - 3x3 + 12x
= 2x3 + 16x + x3 - 3x3 + 12x
= 28x
\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
pt thành nhân tử là ra
Bài 2:
a: \(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y-4\right)\left(x+y+3\right)\)
b: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
c: \(4x^4-32x^2+1\)
\(=4x^4+4x^2+1-36x^2\)
\(=\left(2x^2+1\right)^2-36x^2\)
\(=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
d: \(=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
Bài 1:
a, \(\left(2x+2\right)\left(2x-2\right)=4x^2-4\)
b, đề thiếu
Bài 2:
a, \(\left(x-1\right)\left(x^2+2x+4\right)=\left(x-1\right)^3\)
Thay x = -1 vào đa thức trên, ta được:
\(\left(x-1\right)^3=\left(-1-1\right)^3=-2^3=-8\)
b, \(x^2-2xy-9z^2+y^2=\left(x^2-2xy+y^2\right)-\left(3z\right)^2=\left(x-y\right)^2-\left(3z\right)^2=\left(x-y+3z\right)\left(x-y-3z\right)\)
Thay x = 6; y = -4; z = 20 vào đa thức, ta được:
\(\left(x-y+3z\right)\left(x-y-3z\right)=\left(6+4+3.20\right)\left(6+4-3.20\right)=70.\left(-50\right)=-3500\)
Bài 3:
a, \(x^3-2x^2+x=x^3-x^2-x^2+x=x^2\left(x-1\right)-x\left(x-1\right)=\left(x-1\right)\left(x^2-x\right)=x\left(x-1\right)^2\)
b, \(x+y-y^2-xy=\left(x+y\right)-x\left(y+x\right)=\left(x+y\right)\left(1-x\right)\)
1.
a)\(\left(2x+2\right)\left(2x-2\right)=4x^2-4\)
b) đề thiếu
c) đặt tính ra
2.
a)\(\left(x-1\right)\left(x^2+2x+4\right)=x^3-1\)
Giá trị của biểu thức trên tại x=-1 là:
\(\left(-1\right)^3-1=-2\)
b)\(x^2-2xy-9z^2+y^2=\left(x-y\right)^2-\left(3z\right)^2=\left(x-y+3z\right)\left(x-y-3z\right)\)
Giá trị của biểu thức trên tại x=6; y=-4 và z=20 là:
\(\left[6-\left(-4\right)+3.20\right]\left[6-\left(-4\right)-3.20\right]=\left(10+60\right)\left(10-60\right)=70.\left(-50\right)=-3500\)
3.
a)\(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
b)\(x+y-y^2-xy=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)
bài 1 : ta có : \(A=27x^3+27x^2y+9xy^2+y^3=\left(3x+y\right)^3\)
\(=\left(3.\left(-3\right)+5\right)^3=\left(-9+5\right)^3=\left(-4\right)^3=-64\)
bài 2 : a) ta có : \(12a^2-3ab+8ac-2bc=3a\left(4a-b\right)+2c\left(4a-b\right)\)
\(=\left(3a+2c\right)\left(4a-b\right)\) câu này mk sữa đề lại chút .
b) ta có : câu này đề sai rồi .
nếu phân tích ra nó sẽ thành : \(17x^2+34x-5=\left(17x+17-\sqrt{374}\right)\left(x+\dfrac{17+\sqrt{374}}{17}\right)\)
c) ta có : \(4x^4+81=\left(2x^2\right)^2+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)
câu 3 : a) ta có : \(-3x^2+2x+1=0\Leftrightarrow-3x^2+3x-x+1=0\)
\(\Leftrightarrow-3x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(-3x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-1}{3};x=1\)
b) ta có : \(x\left(x-3\right)=2x-6=x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
vậy \(x=2;x=3\)