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Có x+y+z=0
<=>(x+y+z)+(x+y+z)=0
<=>x+y+z+x+y+z=0
<=>2x+2y+2z=0
<=>(2x+2y+2z).2=0(1)
Tương tự có :(4x+4y+4z).2=0(2)
Từ (1)và(2) có (x2+y2+z2).2=2.(x4+y4+z4)
Chúc bạn học tốt nha
a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
Ta có:
\(x^4\ge0\); \(y^4\ge0\) ;\(z^4\ge0\)
\(\Rightarrow x^4+y^4+z^4\ge0\)
Ta cũng có:
\(x^2\ge0\); \(y^2\ge0\) ;\(z^2\ge0\)
\(\Rightarrow x^2+y^2+z^2\ge0\)
Mà: \(x^4>x^2;y^4>x^2;z^4>z^2\)
\(\Rightarrow x^4+y^4+z^4\ge\left(x^2+y^2+z^2\right):3\) (đpcm)
Bài 3:
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)
\(=\left(x^2-9\right)\left(x^2-1\right)+15\)
\(=x^4-10x^2+9+15\)
\(=x^4-10x^2+24\)
\(=\left(x^2-4\right)\left(x^2-6\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)
a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)
c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)
\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)
1: Phân tích thành nhân tử
c) Ta có: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Lời giải:
$x^3+y^3=(x+y)^3-3xy(x+y)=2^3-3xy.2=8-6xy$
$=8-3.2xy=8-3[(x+y)^2-(x^2+y^2)]=8-3(2^2-34)=98$
----------------
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=34^2-\frac{1}{2}(2xy)^2$
$=34^2-\frac{1}{2}[(x+y)^2-(x^2+y^2)]^2=34^2-\frac{1}{2}(2^2-34)^2=706$
a, \(8^3yz+12^2yz+6xyz+yz\)
\(=512yz+144yz+6xyz+yz\)
\(=yz\left(512+14+6x+1\right)\)
\(=yz\left(527+6x\right)\)
$---$
b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)
\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)
\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)
\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)
$---$
c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)
\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)
$---$
d, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=125\)
b:\(B=x^4+y^4\)
\(=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=125^2-2\cdot2500\)
=10625
c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)
\(C=x^2-y^2=\left(x-y\right)\left(x+y\right)=15\cdot5=75\)
\(x+y+z=0\)
\(\Leftrightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Leftrightarrow2+2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow2\left(xy+yz+xz\right)=-2\)
\(\Leftrightarrow xy+yz+xz=-1\)
\(\Leftrightarrow\left(xy+yz+xz\right)^2=1\)
\(\Leftrightarrow x^2y^2+y^2z^2+x^2z^2+2xyyz+2xyxz+2yzxz=1\)
\(\Leftrightarrow x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)=1\)
\(\Leftrightarrow x^2y^2+y^2z^2+x^2z^2+2xyz\cdot0=1\)
\(\Leftrightarrow x^2y^2+y^2z^2+x^2z^2=1\)(*)
Ta lại có : \(x^2+y^2+z^2=2\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=2^2\)
\(\Leftrightarrow x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2=4\)
\(\Leftrightarrow x^4+y^4+z^4+2\left(x^2y^2+x^2z^2+y^2z^2\right)=4\)
Thay (*) vào đẳng thức ta có :
\(x^4+y^4+z^4+2\cdot1=4\)
\(\Leftrightarrow x^4+y^4+z^4=4-2=2\)
Vậy \(x^4+y^4+z^4=2\)tại \(x+y+z=0;x^2+y^2+z^2=2\)