Cho ( 1 / xy ) + ( 1 / yz ) + ( 1 / xz ) = 0
CMR : A = Căn của x2 / yz ( 1 + x2 ) + Căn của y2 / xz ( 1 + y2 ) + Căn của z2 / xy( 1 + z2 ) \(\le\)3 / 2
cho x+y+z=4 xy+xz+xt+yz+yt+zt=1 tìm GTNN của x2+y2+z2+t2
Cho x+y+z=4 xy+xz+xt+yz+yt+zt=1 tìm GTNN của x2+y2+z2+t2
cho x y z là các số thực dương thỏa mãn x + y + z = 3.Tìm GTLN của A= xy/căn(z2+3) + yz/căn(x2+3) + zx/căn(y2+3)
cho x+y+z=a
x2+y2+z2=b
\(\dfrac{1}{\text{x
}}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\)
Tính xy+yz+xz, x3+y3+z3
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=a^2+b\)
\(\Rightarrow xy+yz+xz=\dfrac{a^2+b}{2}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\Rightarrow\dfrac{xy+yz+xz}{xyz}=\dfrac{1}{c}\)
\(\Rightarrow xyz=c\left(xy+yz+xz\right)\)
\(\Rightarrow xyz=\dfrac{\left(a^2+b\right)c}{2}\)
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+xz\right)\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=a\left(b-\dfrac{a^2+b}{2}\right)+3\dfrac{\left(a^2+b\right)c}{2}\)
\(\Rightarrow x^3+y^3+z^3=a\dfrac{\left(b-a^2\right)}{2}+3\dfrac{\left(a^2+b\right)c}{2}\)
x2+y2+z2≥xy+yz+xz
** Lần sau bạn lưu ý ghi đề bài đầy đủ.
Cho $x,y,z$ là các số thực. CMR $x^2+y^2+z^2\geq xy+yz+xz$
----------------------------
Ta có:
BĐT cần cm tương đương với:
$x^2+y^2+z^2-xy-yz-xz\geq 0$
$\Leftrightarrow 2x^2+2y^2+2z^2-2xy-2yz-2xz\geq 0$
$\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2xz+x^2)\geq 0$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$
(luôn đúng với mọi số thực $x,y,z$)
Do đó ta có đpcm
Dấu "=" xảy ra khi $x=y=z$
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc
1) \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^3-x^3y^2+y^2z^3-y^3z^2-z^2x^2\left(z-x\right)\)
\(=\left(y^2z^3-x^3y^2\right)-\left(y^3z^2-x^2y^3\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z^3-x^3\right)-y^3\left(z^2-x^2\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(z^2+zx+x^2\right)-y^3\left(z-x\right)\left(z+x\right)-z^2x^2\left(z-x\right)\)
\(=\left(z-x\right)\left[y^2\left(z^2+zx+x^2\right)-y^3\left(z+x\right)-z^2x^2\right]\)
\(=\left(z-x\right)\left[\left(y^2z^2+xy^2z+x^2y^2\right)-\left(y^3z+xy^3\right)-z^2x^2\right]\)
\(=\left(z-x\right)\left(y^2z^2+xy^2z+x^2y^2-y^3z-xy^3-z^2x^2\right)\)
\(=\left(z-x\right)\left[\left(y^2z^2-y^3z\right)-\left(x^2z^2-x^2y^2\right)+\left(xy^2z-xy^3\right)\right]\)
\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z^2-y^2\right)+xy^2\left(z-y\right)\right]\)
\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z-y\right)\left(z+y\right)+xy^2\left(z-y\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[y^2z-x^2\left(z+y\right)+xy^2\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y^2z-x^2z-x^2y+xy^2\right)\)
\(=\left(z-x\right)\left(z-y\right)\left[\left(y^2z-x^2z\right)-\left(x^2y-xy^2\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[z\left(y^2-x^2\right)-xy\left(x-y\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[z\left(y-x\right)\left(y+x\right)+xy\left(y-x\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left[z\left(y+x\right)+xy\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left(yz+xz+xy\right)\)
2) \(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-y-x+1\right)\)
\(=\left(z-1\right)\left[\left(xy-y\right)-\left(x-1\right)\right]\)
\(=\left(z-1\right)\left[y\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
cho a+b+c=3,a,b,c>=0 tìm max (x2+y2+z2)(xy+yz+xz)2
Lời giải:
Áp dụng BĐT Cô-si:
$(x^2+y^2+z^2)(xy+yz+xz)^2=(x^2+y^2+z^2)(xy+yz+xz)(xy+yz+xz)$
$\leq \left(\frac{x^2+y^2+z^2+xy+yz+xz+xy+yz+xz}{3}\right)^3$
$=\frac{(x+y+z)^6}{27}=\frac{3^6}{27}=27$
Vậy max của biểu thức là $27$ khi $a=b=c=1$
Cho x2 + y2 + z2 = 10. Tính:
P = ( xy + yz + xz)2 + ( x - yz)2 + ( y - xz)2 + ( z - xy)2