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Bài giải
a) \(\dfrac{1}{x+2}=\dfrac{x.\left(x-2\right)}{\left(x+2\right)\left(x-2\right).x}=\dfrac{x^2-2x}{x\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
b) \(x^2+1=\dfrac{x^2+1}{1}=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}\) giữ nguyên.
c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
Lời giải:
\(B=x(x^2+xy+y^2)-y(y^2+xy+y^2)\)
\(=(x-y)(x^2+xy+y^2)=x^3-y^3=10^3-(-1)^3=1000-(-1)=1001\)
\(C=x^4+10x^3+10x^2+10\)
\(=x^4+9x^3+x^3+9x^2+x^2+10\)
\(=x^3(x+9)+x^2(x+9)+x^2+10\)
\(=(x+9)(x^3+x^2)+x^2+10\)
\(=(-9+9)[(-9)^3+(-9)^2]+(-9)^2+10\)
\(=0+(-9)^2+10=91\)
Thay $x=-1$ vào biểu thức:
\(D=x^2(x+y)-xy(x-y)-x(y^2+1)\)
\(=(-1)^2(x+y)-(-1)y(x-y)-(-1)(y^2+1)\)
\(=x+y+y(x-y)+(y^2+1)\)
\(=x+y+xy-y^2+y^2+1=x+y+xy+1\)
\(=(x+1)(y+1)=(-1+1)(y+1)=0\)
\(a,x^2+y^2-x-y=8\)
\(\Rightarrow x^2-x+\frac{1}{4}+y^2-y+\frac{1}{4}-8,5=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-8,5=0\)
Ta có : \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-8,5\ge-8,5\forall x;y\)
Để VP=0 và là các số nguyên
=>\(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=8,5\)
a/ x^2 + y^2 - x - y = 8
<=> 4x^2 + 4y^2 - 4x - 4y = 32
<=> (2x - 1)^2 + (2y - 1)^2 = 34
<=> (2x - 1)^2 = 9 và (2y - 1)^2 = 25
Hoặc (2x - 1)^2 = 25 và (2y - 1)^2 = 9
a/\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\left(đpcm\right)\)
b/ \(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+xy^3-y^4=x^4-y^4\left(đpcm\right)\)
c/ \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+y^2+xy+yz+z^2+zx+yz=x^2+y^2+z^2+2xy+2yz+2zx\left(đpcm\right)\)
d/ \(\left(x+y+z\right)^3=\left(x+y\right)^3+3\left(x+y\right)^2z+3z^2\left(x+y\right)+z^3\)
\(=\left(x+y\right)^3+3z\left(x^2+2xy+y^2\right)+3z^2\left(x+y\right)+z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3x^2z+6xyz+3y^2z+3z^2x+3yz^2+z^3\)
\(=x^3+y^3+z^3+3xyz+3x^2y+3xy^2+3x^2z+3y^2z+3y^2x+3yz^2+3xyz\)
\(=x^3+y^3+z^3+\left(x+z\right)\left(3xy+3xz+3y^2+3yz\right)\)
\(=x^3+y^3+z^3+\left(x+z\right)\left[3x\left(y+z\right)+3y\left(y+z\right)\right]\)
\(=x^3+y^3+z^3+\left(x+z\right)\left(y+z\right)\left(3x+3y\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\) (đpcm)
a, Xét vế trái ta có:
(x-1)(x^2+ x+1)=x^3+ x^2+ x- x^2- x-1
=x^3+ (x^2- x^2)+(x-x)-1
=x^3-1
Vậy...
b,Xét vế trái ta có:(x^3+ x^2y+ xy^2+ y^3)(x-y)
=x^4- x^3y+ x^3y- x^2- y^2+ x^2y^2- xy^3+ xy^3- y^4
=x^4-y^4
Vậy ........
c, Xét vế trái ta có:
(x+y+z)^2=(x+y+z)(x+y+z)
=x^2+ xy+ xz+ yx+y^2+ yz+ zx+ zy+ z^2
=x^2+ y^2+ z^2+ 2xy+ 2xz+ 2yz
Vậy...............
d, Xé vế trái ta có:
(x+y+x)^3=(x+y+z)(x+y+z)(x+y+z)(x+y+z)
=(x^2+y^2+z^2+2xy+2xz+2yz)(x+y+z)
=x^3+ xy^2+ xz^2+ 2x^2y+ 2xyz+ 2x^2z+ x^2y+ y^3+ yz^2+2xy^2+ 2y^2z+z^3+ 2xyz+ x^2z+ y^2z+2xyz+ 2yz^2+ 2xz^2
=x^3+ 3xy^2+ 6xy+ 3x^2y+3xz^2+ 3x^2z+ 3yz^2+ y^3z^3 (1)
Xét vế phải ta có:x^3+ y^3+ z^3+ 3(x+y)(x+y)(y+z)
=x^3+ y^3+ z^3+ 3(xy+ xz+ y^2+ yz)(z+x)
=x^3+ y^3+ z^3+ 3(xyz+ xz^2+ y^2z+ yz^2+ x^2y+ x^2z+ xy^2+xyz)
=x^2+ y^3+ z^3 +3(2xyz+ xz^2+ y^2z+ yz^2+x^2y+x^2z+ xy^2)
=x^3+ y^3+ z^3+6xyz+ 3xz^2+ 3y^2z+3yz^2+ 3x^2y+3x^2z+3xy^2(2)
Từ (1) và (2)=>.......
Đặt \(B=\frac{\left(x+y+1\right)^2}{xy+x+y}\)
Ta có bđt sau \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\) tự chứng mình nha
Áp dụng \(a=x,b=y,c=1\)
Ta có : \(B=\frac{\left(x+y+1\right)^2}{xy+x+y}\ge3\)
Ta có : \(A=\frac{1}{B}+B=\frac{1}{B}+\frac{B}{9}+\frac{8B}{9}\ge\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)
Dấu " = " xảy ra khi \(x=y=1\)
