Cho 1/x + 1/y + 1/z = 3
và 1/x^2 + 1/y^2 + 1/z^2 = 1
CM: x+y+z = 4xyz
Những câu hỏi liên quan
Cho 1/x + 1/y + 1/z = 3
va 1/x^2 + 1/y^2 + 1/z^2 = 1
CM: x+y+z = 4xyz
cho x^2+y^2+z^2=5/2 va x,y,z>0 cm 1/x+1/y<1/xyz+1/z\(cho x^2+y^2+z^2=5/2 va x,y,z>0 cm 1/x+1/y<1/xyz+1/z\)
cm trong 3 so x,y,z ton tai 1 so bang tong 2 so con lai biet
x.(y-z)^2+y.(x-z)^2+z.(x-y)^2-x^3-y^3-z^3+4xyz=0
1/x+1/y+1/z và 1/x^2+1/y^2+1/z^2=1.CMR x+y+z=4xyz
cho x,y.z là ba số # +- 1 sao cho xy +yz +xz =1 . CMR : x/1-x^2 + y/1-y^2 + z/1-z^2 = 4xyz / (1 -x^2 ) . (1- y^2) . ( 1- z^2)
1.Cho x^2+ 4x+1 = 0
Tính A= ( x + 1/x )^2 + (x^2 + 1/x^2 )^2 + ( x^3+ 1/x^3 )^2
2.Cho các số thực x, y khác 0 sao cho x+ 1/y và y+ 1/x là những số nguyên . CMR x^3y^3 + 1/x^3y^3 là số nguyên.
3.Cho x,y,z khác 0 tm x(y+z)^2+y(z+x)^2+z(x+y)^2=4xyz
Cho x,y,z>0; \(x^2+y^2+z^2+4xyz=2\left(xy+yz+zx\right)\).Tìm GTLN của P=x(1-y)(1-z)
\(x^2+y^2+z^2+4xyz=2\left(xy+yz+zx\right)\\ \Leftrightarrow\left(x-y-z\right)^2=\left(1-x\right)4yz\ge0\\ \Leftrightarrow1-x\ge0\Leftrightarrow0< x\le1\\ \Leftrightarrow\left(x-y-z\right)^2=\left(1-x\right)4yz\le\left(1-x\right)\left(y+z\right)^2\\ \Leftrightarrow x^2-2x\left(y+z\right)+\left(y+z\right)^2\le\left(1-x\right)\left(y+z\right)^2\\ \Leftrightarrow x^2-2x\left(y+z\right)\le\left(y+z\right)^2\left(1-x-1\right)=-x\left(y+z\right)^2\\ \Leftrightarrow x-2\left(y+z\right)\le-\left(y+z\right)^2\\ \Leftrightarrow x\le\left(y+z\right)\left[2-\left(y+z\right)\right]\)
Đặt \(2-\left(y+z\right)=t\)
\(P=x\left(1-y\right)\left(1-z\right)\le x\left(\dfrac{1-y+1-z}{2}\right)^2=\dfrac{x\left[2-\left(y+z\right)\right]^2}{4}\\ \Leftrightarrow4P\le x\left[2-\left(y+z\right)\right]^2\le\left(y+z\right)\left[2-\left(y+z\right)\right]^3\\ \Leftrightarrow4P\le t^3\left(2-t\right)=\dfrac{27}{16}-\dfrac{\left(4t^2+4t+3\right)\left(2t-3\right)^2}{16}\)
Mà \(-\dfrac{\left(4t^2+4t+3\right)\left(2t-3\right)^2}{16}\le0\Leftrightarrow4P\le\dfrac{27}{16}\Leftrightarrow P\le\dfrac{27}{64}\)
Dấu \("="\Leftrightarrow x=\dfrac{3}{4};y=z=\dfrac{1}{4}\)
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Phân tích các đa thức sau thành nhân tử
a,2x2+3xy-14y2
b,(x-7)(x-5)(x-3)(x-1)+7
c,(x-3)2+(x-3)(3x-1)-2(3x-1)2
d,xy(x-y)+yz(y-z)+zx(z-x)
f,x(y+z)2+y(z+x)2+z(x+y)2-4xyz
a: \(2x^2+3xy-14y^2\)
\(=2x^2+7xy-4xy-14y^2\)
\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)
\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)
\(=\left(2x+7y\right)\left(x-2y\right)\)
b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)
\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)
\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)
\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)
\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)
c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)
\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)
\(=\left(7x-5\right)\left(-2x-2\right)\)
\(=-2\left(x+1\right)\left(7x-5\right)\)
d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)
\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)
\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)
\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)
\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)
\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)
\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
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cho x,y,z \(\ne\)-1,1 ,xy+zx+zy=1
\(\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\)