Rút gọn biểu thức: \(\left( {3{x^2} - 5xy - 4{y^2}} \right).\left( {2{x^2} + {y^2}} \right) + \left( {2{x^4}y - {x^3}{y^3} - {x^2}{y^4}} \right):\left( {\dfrac{1}{5}xy} \right)\)
Rút gọn rồi tính giá trị của biểu thức khi x=1;y=\(-3\frac{1}{4}\)
\(\frac{\left(x-y\right)^2+xy}{\left(x+y\right)^2-xy}\)\(\left[1:\frac{x^5+y^5+x^3y^2+x^2y^3}{\left(x^3y^3\right)\left(x^3+y^3+x^2y+xy^2\right)}\right]\)
Rút gọn biểu thức
a. Q= \(\left(x-y\right)^2\)-4(x-y)(x+2y)+4\(\left(x+2y\right)^2\)
b. A=\(\left(xy+2\right)^3\)-6\(\left(xy+2\right)^2\)+12(xy+2)-8
c. \(\left(x+2\right)^3\)+\(\left(x-2\right)^3\)-2x(\(x^2\)+12)
a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)
\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)
\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)
\(Q=\left(x-y-2x-4y\right)^2\)
\(Q=\left(-x-5y\right)^2\)
b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)
\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)
\(A=\left[\left(xy+2\right)-2\right]^3\)
\(A=\left(xy+2-2\right)^3\)
\(A=\left(xy\right)^3\)
\(A=x^3y^3\)
c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)
\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)
\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)
\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)
\(=0\)
a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2
=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2
b: =(xy+2-2)^3=(xy)^3=x^3y^3
c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)
=24x+2x^3-2x^3-24x
=0
Tìm tập xác định, rồi rút gọn biểu thức:
B = \(\dfrac{y-x}{xy}\) : [\(\dfrac{y^2}{\left(x-y\right)^2\left(x+y\right)}\) - \(\dfrac{2x^2y}{x^4-2x^2y^2+y^4}\) + \(\dfrac{x^2}{\left(y^2-x^2\right)\left(x+y\right)}\)]
Tính giá trị của B với x = -\(\dfrac{1}{2}\), y = 2
Rút gọn biểu thức:
\(a)\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\)
b) \(\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\)
c) \(\left( {4{\rm{x}} - 1} \right)\left( {6y + 1} \right) - 3{\rm{x}}\left( {8y + \dfrac{4}{3}} \right)\)
d) \(\left( {x + y} \right)\left( {x - y} \right) + \left( {x{y^4} - {x^3}{y^2}} \right):\left( {x{y^2}} \right)\)
a)
\(\begin{array}{l}\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\\ = x.{x^2} + x.xy + x.{y^2} - y.{x^2} - y.xy - y.{y^2}\\ = {x^3} + {x^2}y + x{y^2} - {x^2}y - x{y^2} - {y^3}\\ = {x^3} - {y^3}\end{array}\)
b)
\(\begin{array}{l}\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\ = x.{x^2} + x.\left( { - xy} \right) + x{y^2} + y.{x^2} + y.\left( { - xy} \right) + y.{y^2}\\ = {x^3} - {x^2}y + x{y^2} + {x^2}y - x{y^2} + {y^3}\\ = {x^3} + {y^3}\end{array}\)
c)
\(\begin{array}{l}\left( {4{\rm{x}} - 1} \right)\left( {6y + 1} \right) - 3{\rm{x}}\left( {8y + \dfrac{4}{3}} \right)\\ = 4{\rm{x}}.6y + 4{\rm{x}}.1 - 1.6y - 1.1 - 3{\rm{x}}.8y - 3{\rm{x}}.\dfrac{4}{3}\\ = 24{\rm{x}}y + 4{\rm{x}} - 6y - 1 - 24{\rm{x}}y - 4{\rm{x}}\\ = - 6y - 1\end{array}\)
d)
\(\begin{array}{l}\left( {x + y} \right)\left( {x - y} \right) + \left( {x{y^4} - {x^3}{y^2}} \right):\left( {x{y^2}} \right)\\ = x.x + x.\left( { - y} \right) + y.x + y.\left( { - y} \right) + \left( {x{y^4}} \right):\left( {x{y^2}} \right) + \left( { - {x^3}{y^2}} \right):\left( {x{y^2}} \right)\\ = {x^2} - xy + xy - {y^2} + {y^2} - x^2\\ = 0\end{array}\)
Cho x,y là hai số trái dấu và x+y=1
a) Rút gọn biểu thức A=\(\dfrac{y-x}{xy}:\left[\dfrac{y^2}{\left(x-y\right)^2}-\dfrac{2x^2y}{\left(x^2-y^2\right)^2}+\dfrac{x^2}{y^2-x^2}\right]\)
b) CM: A<-4
Rút gọn biểu thức:
a) \(A=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
b) \(B=3x^2\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x^4+x^2+1\right)+\left(x^2-1\right)^3\)
c) \(C=\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)
d) \(D=\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
Rút gọn các biểu thức sau:
a) A = \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
b) B = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
c) C = \(\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2+\sqrt{x}}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
d) D = \(\sqrt{\dfrac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\dfrac{a+x^2}{x}+2\sqrt{a}}\) với a > 0, x > 0.
Rút gọn:
\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^2}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(=\frac{x^2+xy+y^2}{x+y}.\left(\frac{1}{\left(x-y\right)x}-\frac{3y^2}{x\left(x^3-y^3\right)}-\frac{y}{x\left(x^2+xy+y^2\right)}\right)\)
\(=\frac{x^2+xy+y^2}{x+y}.\frac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^2-y^2}{x\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}=\frac{1}{x}\)
Rút gọn các phân thức sau:
a) \(\dfrac{6x^2y^2}{8xy^{ }5}\)
b) \(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
c) \(\dfrac{2x^2+2x
}{x+1}\)
d) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
e) \(\dfrac{36\left(x-2\right)^3}{32-16x}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)