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Có \(\left(x+y+z\right)^3\)
\(=\left[\left(x+y\right)+z\right]^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3\left(x+y\right)\left[\left(x+y\right)z+z^2\right]+z^3\)
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3\left(x+y\right)\left(xz+yz+z^2\right)\)
=\(x^3+y^3+z^3+3\left(x+y\right)\left(xz+xy+yz+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Ta có: (x + y + z)3 = x3 + y3 + z3 + 3(x + y)(y + z)(z + x)
\(\Leftrightarrow\) (x + y + z)3 - x3 - y3 - z3 = 3(x + y)(y + z)(z + x)
Phân tích VT ta được:
(x + y + z)3 - x3 - y3 - z3 = \(\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
= (x + y)3 + z3 + 3z(x + y)(x + y + z) - x3 - y3 - z3
= x3 + y3 + 3xy(x + y) + z3 + 3z(x + y)(x + y + z) - x3 - y3 - z3
= 3xy(x + y) + 3z(x + y)(x + y + z)
= 3(x +y)(xy + xz + yz + z2)
= 3(x +y)\(\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
= 3(x + y)(y + z)(z + x) (đpcm)
Bài này cần áp dụng công thức (x + y)3 = x3 + y3 + 3xy(x + y) nhiều lần để phân tích nhé bạn.
a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Ta có: \(-x^2+2x-3=-x^2+2x-1-2=-\left(x-1\right)^2-2\le-2\) (1)
Và \(A=\dfrac{-5}{x^2-2x+3}=\dfrac{5}{-x^2+2x+3}\) (2)
Từ (1);(2)\(\Rightarrow A\ge-\dfrac{5}{2}\) Vậy min A=-5/2 khi x=1
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
= \(\left[\left(x+y+z\right)-\left(x+y\right)\right]^2\)
= \(z^2\)
Ta có:(x + y + z)2 - 2(x + y + z) (x + y) + (x + y)2
=[(x+y+z)-(x+y)]2=z2
Vì x,y,z >0
Áp dụng BĐT Cosy:
\(x+y\ge2\sqrt{xy}\) (1)
\(x+z\ge2\sqrt{xz}\) (2)
\(y+z\ge2\sqrt{yz}\) (3)
Nhân 3 vế ta được:
\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{xz}\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8xyz\left(đpcm\right)\)
Chúc bạn học tốt
áp dụng BĐT cauchy, ta có:
\(\left[{}\begin{matrix}x+y\ge2\sqrt{xy}\\y+z\ge2\sqrt{yz}\\x+z\ge2\sqrt{xz}\end{matrix}\right.\)
nhân vế theo vế các BĐT trên, ta được:
\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{xz}=8\sqrt{\left(xyz\right)^2}=8xyz\)