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1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
1, \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(A=5x^3-15x+7x^2-5x^3-7x^2\)
\(A=\left(5x^3-5x^3\right)+\left(7x^2-7x^2\right)-15x\)
\(A=-15x\)
Thay \(x=-5\) vào A ta được:
\(-15\cdot-5=75\)
Vậy: ....
2. \(B=x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(B=x^3-3x+7x^2-5x^3-7x^2\)
\(B=\left(x^3-5x^3\right)+\left(7x^2-7x^2\right)-3x\)
\(B=-4x^3-3x\)
Thay \(x=10,y=-1\) vào B ta được:
\(-4\cdot10^3-3\cdot10=-4\cdot1000-3\cdot10=-4000-30=-4030\)
Vậy: ....
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
a)
\(\begin{array}{l}A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\\A = x - 0,2 - \dfrac{1}{3}x - 2 + 2 - \dfrac{2}{3}x\\ = \left( {x - \dfrac{1}{3}x - \dfrac{2}{3}x} \right) + \left( {\dfrac{{ - 1}}{2} - 2 + 2} \right)\\ = - \dfrac{1}{2}\end{array}\)
Vậy \(A = - \dfrac{1}{2}\) không phụ thuộc vào biến x
b)
\(\begin{array}{l}B = \left( {x - 2y} \right)\left( {{x^2} + 2{\rm{x}}y + 4{y^2}} \right) - \left( {{x^3} - 8{y^3} + 10} \right)\\B = \left[ {x - {{\left( {2y} \right)}^3}} \right] - {x^3} + 8{y^3} - 10\\B = {x^3} - 8{y^3} - {x^3} + 8{y^3} - 10 = - 10\end{array}\)
Vậy B = -10 không phụ thuộc vào biến x, y.
c)
\(\begin{array}{l}C = 4{\left( {x + 1} \right)^2} + {\left( {2{\rm{x}} - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) - 4{\rm{x}}\\{\rm{C = 4}}\left( {{x^2} + 2{\rm{x}} + 1} \right) + \left( {4{{\rm{x}}^2} - 4{\rm{x}} + 1} \right) - 8\left( {{x^2} - 1} \right) - 4{\rm{x}}\\C = 4{{\rm{x}}^2} + 8{\rm{x}} + 4 + 4{{\rm{x}}^2} - 4{\rm{x}} + 1 - 8{{\rm{x}}^2} + 8 - 4{\rm{x}}\\C = \left( {4{{\rm{x}}^2} + 4{{\rm{x}}^2} - 8{{\rm{x}}^2}} \right) + \left( {8{\rm{x}} - 4{\rm{x}} - 4{\rm{x}}} \right) + \left( {4 + 1 + 8} \right)\\C = 13\end{array}\)
Vậy C = 13 không phụ thuộc vào biến x
a) x(x - y) + y (x + y) = x2 – xy +yx + y2= x2+ y2
với x = -6, y = 8 biểu thức có giá trị là (-6)2 + 82 = 36 + 64 = 100
b) x(x2 - y) - x2 (x + y) + y (x2– x) = x3 – xy – x3 – x2y + yx2 - yx
= -2xy
Với x = \(\dfrac{1}{2}\), y = -100 biểu thức có giá trị là -2 . \(\dfrac{1}{2}\) . (-100) = 100.
a)x(x-y)+y(x+y)=x2-xy+xy+y2=x2+y2
Tại x=-6 y=8 ta được :
(-6)2+82=36+64=100
b) x(x2-y)-x2(x+y)+y(x2-x)
=x3-xy-x3-x2y+x2y-xy=-2xy
Tại x=\(\dfrac{1}{2}\) y=-100 ta được :
(-2).\(\dfrac{1}{2}\).(-100)=-1.-100=100
\(3,x=\dfrac{1}{2},y=-1\)
\(\Rightarrow C=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+1\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-1\right)-1\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)
\(\Rightarrow C=\dfrac{1}{2}\left(\dfrac{1}{4}+1\right)-\dfrac{1}{4}\left(-\dfrac{1}{2}\right)-\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)
\(\Rightarrow C=\dfrac{1}{2}.\dfrac{5}{4}+\dfrac{1}{8}-\left(-\dfrac{1}{4}\right)\)
\(\Rightarrow C=\dfrac{5}{8}+\dfrac{1}{8}+\dfrac{1}{4}\)
\(\Rightarrow C=1\)
\(4,x=\dfrac{1}{2},y=-100\)
\(\Rightarrow D=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+100\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-100\right)-100\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)
\(\Rightarrow D=\dfrac{1}{2}\left(\dfrac{1}{4}+100\right)-\dfrac{1}{4}\left(-\dfrac{199}{2}\right)-100\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)
\(\Rightarrow D=\dfrac{1}{2}.\dfrac{401}{4}+\dfrac{199}{8}-100.\left(-\dfrac{1}{4}\right)\)
\(\Rightarrow D=\dfrac{401}{8}+\dfrac{199}{8}+25\)
\(\Rightarrow D=100\)
3: C=x^3-xy-x^3-x^2y+x^2y-xy
=-2xy=-2*1/2*(-1)=1
4: D=x^3-xy-x^3-x^2y+x^2y-xy
=-2xy
=-2*1/2*(-100)=100