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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
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
\(3,x=\dfrac{1}{2},y=-1\)
\(\Rightarrow C=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+1\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-1\right)-1\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)
\(\Rightarrow C=\dfrac{1}{2}\left(\dfrac{1}{4}+1\right)-\dfrac{1}{4}\left(-\dfrac{1}{2}\right)-\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)
\(\Rightarrow C=\dfrac{1}{2}.\dfrac{5}{4}+\dfrac{1}{8}-\left(-\dfrac{1}{4}\right)\)
\(\Rightarrow C=\dfrac{5}{8}+\dfrac{1}{8}+\dfrac{1}{4}\)
\(\Rightarrow C=1\)
\(4,x=\dfrac{1}{2},y=-100\)
\(\Rightarrow D=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+100\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-100\right)-100\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)
\(\Rightarrow D=\dfrac{1}{2}\left(\dfrac{1}{4}+100\right)-\dfrac{1}{4}\left(-\dfrac{199}{2}\right)-100\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)
\(\Rightarrow D=\dfrac{1}{2}.\dfrac{401}{4}+\dfrac{199}{8}-100.\left(-\dfrac{1}{4}\right)\)
\(\Rightarrow D=\dfrac{401}{8}+\dfrac{199}{8}+25\)
\(\Rightarrow D=100\)
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