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a) 6x2 - 12x
= 6x(x - 2)
b) x2 + 2x + 1 - y2
= (x2 + 2x + 1) - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
c) x + y + z + x2 + xy + xz
= (x + x2) + (y + xy) + (z + xz)
= x(1 + x) + y(1 + x) + z(1 + x)
= (x + y + z)(x + 1)
d) xy + xz + y2 + yz
= (xy + xz) + (y2 + yz)
= x(y + z) + y(y + z)
= (x + y)(x + z)
e) x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x2 + 1)(x + 1)
f) xy + y - 2x - 2
= (xy + y) - (2x + 2)
= y(x + 1) - 2(x + 1)
= (y - 2)(x + 1)
g) x3 + 3x - 3x2 - 9
= (x3 - 3x2) + (3x - 9)
= x2(x - 3) + 3(x - 3)
= (x2 + 3)(x - 3)
h) x2 - y2 - 2x - 2y
= (x2 - y2) - (2x + 2y)
= (x + y)(x - y) - 2(x + y)
= (x + y)(x - y - 2)
i) 7x2 - 7xy - 5x = 5y
mk thấy con này sai sai ý
a)\(^{ }\left(-2\right)^5:\left(-2\right)^3=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)
b)\(\left(-y\right)^7:\left(-y\right)^3=\left(-y\right)^{7-3}=\left(-y\right)^4\)
c)\(x^{12}:\left(-x^{10}\right)=-\left(x^{12}\right):\left(x^{10}\right)=-\left(x^{12-10}\right)=-\left(x^2\right)\)
d)\(\left(2x^6\right):\left(2x\right)^3=2\left(x^6\right):8\left(x^3\right)=\frac{2}{8}x^{6-3}=\frac{1}{4}x^3\)
e)\(\left(-3x\right)^5;\left(-3x\right)^3=\left(-3x\right)^{5-3}=\left(-3x\right)^2\)
f)\(\left(xy^2\right)^4:\left(xy^2\right)^2=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2\)
Bài làm
a) \(\left(-2\right)^5:\left(-2\right)^3\)
\(=\left(-2\right)^{5-3}\)
\(=\left(-2\right)^2\)
\(=4\)
b) \(\left(-y\right)^7:\left(-y\right)^3=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)
c) \(x^{12}:\left(-x\right)^{10}=x^{12}:x^{10}=x^{12-10}=x^2\)
d) \(\left(2x\right)^6:\left(2x\right)^3=\left(2x\right)^{6-3}=\left(2x\right)^3=8x^3\)
e) \(\left(-3x\right)^5:\left(-3x\right)^2=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)
f) \(\left(xy^2\right)^4:\left(xy^2\right)^2=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)
# Học tốt #
a) (-2)5 : (-2)3 = (-2)2
b) (-y)7 : (-y)3 = (-y)4
c) x12 : (-x10) = (-x)12
d) 2x6 : 2x3 = 2x3
e) (-3x)5 : (-3x)2 = (-3x)3
f) (xy2)4 : (xy2)2 = (xy2)2
\(a,\dfrac{1}{3x-3y}=\dfrac{x-y}{3\left(x-y\right)^2};\dfrac{1}{x^2-2xy+y^2}=\dfrac{3}{3\left(x-y\right)^2}\\ b,\dfrac{3}{x^2-3x}=\dfrac{6}{2x\left(x-3\right)};\dfrac{5}{2x-6}=\dfrac{5x}{2x\left(x-3\right)}\\ c,\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{3-x}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{x^2-9}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}\)
\(d,\dfrac{1}{x^2+xy}=\dfrac{xy-y^2}{xy\left(x+y\right)\left(x-y\right)};\dfrac{1}{xy-y^2}=\dfrac{x^2+xy}{xy\left(x-y\right)\left(x+y\right)};\dfrac{2}{y^2-x^2}=\dfrac{-2xy}{xy\left(x-y\right)\left(x+y\right)}\)
a)-(x-y)(x2+xy-1)=-(x3+x2y-x-x2y-xy2+y)
=-(x3-xy2-x+y)
=-x3+xy2+x-y
b)x2(x-1)-(x3+1)(x-y)=x3-x2-x3+x2y-x+y
=-x2+x2y-x+y
c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x2-3x-4x+2-15x2-10x-3x-2
=-9x2-20x
d) hình như bạn ghi lỗi
Bài 2: C=x(x2-y)-x2(x+y)+y(x2-x)
=x3-xy-x3-x2y+x2y-xy
=-2xy
Thay x=1/2,y=-1 vào C, ta có:
C=-2.1/2.(-1)=1
Vậy C=1 khi x=1/2 và y=-1.
\(A=\left(3x+5\right)\left(2x-1\right)-\left(1-4x\right)\left(3x+2\right)\)
\(=6x^2+7x-5+12x^2+5x-2\)
\(=18x^2+12x-7\)
\(\left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Thay \(x=-2\) vào biểu thức A ta được :
\(A=18\left(-2\right)^2+12\left(-2\right)-7=41\)
Thay \(x=2\) vào biểu thức A ta được :
\(A=18.2^2+12.2-7=89\)
\(B=\left(2x+y\right)\left(2x-y\right)+xy\left(x-y\right)-xy\left(x+y\right)\)
\(=4x^2-y^2+x^2y-xy^2-x^2y-xy^2\)
\(=4x^2-2xy^2-y^2\)
Thay \(x=0\) và \(y=-1\) vào biểu thức B ta được :
\(B=4.0^2-2.0.\left(-1\right)^2-\left(-1\right)^2=-1\)
1: \(=-3x^3y\cdot2x^2y^3+3x^3y\cdot xy^2+3x^3y\cdot\dfrac{1}{3}\cdot5\)
\(=-6x^5y^4+3x^4y^3+5x^3y\)
2: \(=\dfrac{1}{3}x\cdot3x-\dfrac{1}{3}x\cdot6+2\cdot3x-6\cdot2\)
\(=x^2-2x+6x-12=x^2+4x-12\)