Cho x,y,z > 0. c/m
x3/yz+y3/xz+z3/yx >= x+y+z
Mình đang cần gấp! Giúp mình với ạ
Bài 3: Chứng minh rằng:
a) (x+y+z)2= x2+y2+z2+2xy+2xz+2yz
b) (x-y).(x2+y2+z2-xy-yz-xz)= x3+y3+z3-3xyz
c) (x+y+z)3= x3+y3+z3+3.(x+y).(y+z).(z+x)
Bài 3:
a, (\(x\)+y+z)2
=((\(x\)+y) +z)2
= (\(x\) + y)2 + 2(\(x\) + y)z + z2
= \(x^2\) + 2\(xy\) + y2 + 2\(xz\) + 2yz + z2
=\(x^2\) + y2 + z2 + 2\(xy\) + 2\(xz\) + 2yz
b, (\(x-y\))(\(x^2\) + y2 + z2 - \(xy\) - yz - \(xz\))
= \(x^3\) + \(xy^2\) + \(xz^2\) - \(x^2\)y - \(xyz\) - \(x^2\)z - y3
Đến dây ta thấy xuất hiện \(x^3\) - y3 khác với đề bài, em xem lại đề bài nhé
c,
(\(x\) + y + z)3
=(\(x\) + y)3 + 3(\(x\) + y)2z + 3(\(x\)+y)z2 + z3
= \(x^3\) + 3\(x^2\)y + 3\(xy^{2^{ }}\) + y3 + 3(\(x\)+y)z(\(x\) + y + z) + z3
= \(x^3\) + y3 + z3 + 3\(xy\)(\(x\) + y) + 3(\(x+y\))z(\(x+y+z\))
= \(x^3\) + y3 + z3 + 3(\(x\) + y)( \(xy\) + z\(x\) + yz + z2)
= \(x^3\) + y3 + z3 + 3(\(x\) + y){(\(xy+xz\)) + (yz + z2)}
= \(x^3\) + y3 + z3 + 3(\(x\) + y){ \(x\)( y +z) + z(y+z)}
= \(x^3\) + y3 + z3 + 3(\(x\) + y)(y+z)(\(x+z\)) (đpcm)
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
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=a^2+b\)
\(\Rightarrow xy+yz+xz=\dfrac{a^2+b}{2}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\Rightarrow\dfrac{xy+yz+xz}{xyz}=\dfrac{1}{c}\)
\(\Rightarrow xyz=c\left(xy+yz+xz\right)\)
\(\Rightarrow xyz=\dfrac{\left(a^2+b\right)c}{2}\)
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+xz\right)\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=a\left(b-\dfrac{a^2+b}{2}\right)+3\dfrac{\left(a^2+b\right)c}{2}\)
\(\Rightarrow x^3+y^3+z^3=a\dfrac{\left(b-a^2\right)}{2}+3\dfrac{\left(a^2+b\right)c}{2}\)
CMR: x3+y3+z3-3xyz= (x+y+z)(x2+y2+z2- xy - yz - xz)
Ta có: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-\left[3xy\left(x+y+z\right)\right]\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-zy+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)(đpcm)
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 x,y,z >0 thỏa mãn xy+yz+zx=673
CMR: \(\frac{x}{x^2-yz+2019}+\frac{y}{y^2-xz+2019}+\frac{z}{z^2-yx+2019}\ge\frac{1}{x+y+z}\)
Đk: $x\geq \frac{1}{2}$
Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$
$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$
$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$
$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$
Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$
$\Rightarrow $ Pt $(*)$ vô nghiệm
Cho các số dương x,y,z thỏa mãn điều kiện xy + yz + xz =671
Cmr \(\frac{x}{x^2-yz-2013}+\frac{y}{y^2-xz-2013}+\frac{z}{z^2-yx-2013}\ge\frac{1}{x+y+z}\)
Cho các số dương x,y,z thỏa mãn điều kiện xy + yz + xz =671
Cmr \(\frac{x}{x^2-yz-2013}+\frac{y}{y^2-xz-2013}+\frac{z}{z^2-yx-2013}\ge\frac{1}{x+y+z}\)
\(VT=\frac{x^2}{x^3-xyz-2013x}+\frac{y^2}{y^3-xyz-2013y}+\frac{z^2}{z^3-xyz-2013z}\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013\left(x+y+z\right)}\)
\(=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\right]}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)=VP
đúng rồi ạ nhưng chỉ cần c/m đẳng thức phụ như thế này thôi ạ\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\) =>\(\frac{\left(a+b\right)2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) hay \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) là xong
(\sqrt((x+yz)(y+xz)))/(xy+z)+(\sqrt((y+xz)(z+xy)))/(x+yz)+(\sqrt((x+yz)(z+xy)))/(y+xz)
Với x,y,z>0 thỏa mãn x+y+z=1