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a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)
c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)
\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
a )
Sử dụng Cô-si , ta có :
\(x+y\ge2\sqrt{xy}\) (1)
\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\) (2)
Nhân cả vế (1) vế (2) lại ta có :
\(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge2\sqrt{xy}.2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}=4\)
\(\LeftrightarrowĐPCM.\)
+/ nếu a,b,c>0 hoặc 2 số âm và 1 số dương (abc>0)thì:
M=1+1+1+1=4
+/ nếu a,b,c<0 hoặc 1 số âm và 2 số dương(abc<0) thì:
M=1-1+1-1=0
Sửa lại đề nha: x+y+z=0
a)
Xét x+y+z=0
(x+y+z)2=02
x2+y2+z2+2xy+2yz+2zx=0
=> x2+y2+z2=-2xy-2yz-2zx
Xét \(\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
= \(\dfrac{x^2+y^2+z^2}{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)}\)
=\(\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\)
=\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2-2xy-2yz-2zx}\)(1)
Thay x2+y2+z2=-2xy-2yz-2zx vào (1)
=>\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2+x^2+y^2+z^2}\\=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2}\\ =\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\\ =\dfrac{1}{3}\)
b)
Xét x+y+z=0 ba lần:
- Lần 1:x+y+z=0
<=> x+y=0-z
<=>(x+y)2=(0-z)2
<=>x2+2xy+y2=z2
<=>x2+y2-z2=-2xy(1)
-Lần 2: x+y+z=0
<=> y+z=0-x
<=>(y+z)2=(0-x)2
<=>y2+2yz+z2=x2
<=>y2+z2-x2=-2yz(2)
-Lần 3: x+y+z=0
<=>z+x=0-y
<=>(z+x)2=(0-y)2
<=>z2+2zx+x2=y2
<=> z2+x2-y2=-2zx(3)
Thay (1),(2),(3) vào Q, ta có:
=>\(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}=\dfrac{\left(-2xy\right)\left(-2yz\right)\left(-2zx\right)}{16xyz}\\=\dfrac{\left(-2yz\right)\left(-2zx\right)}{-8z}\\ =\dfrac{y\left(-2zx\right)}{4}\\ =\dfrac{-2xyz}{4}\\ =-\dfrac{xyz}{2}\)
\(a,\dfrac{1}{x^2-x}+\dfrac{2x}{4x^3}-\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{2x^2}-\dfrac{1}{x^2+x+1}\)
\(=\dfrac{2x\left(x^2+x+1\right)+\left(x-1\right).\left(x^2+x+1\right)-2x^2.\left(x-1\right)}{2x^2.\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\dfrac{2x^3+2x^2+2x+x^3-1-2x^3+2x^2}{2x^2.\left(x^3-1\right)}\)
\(=\dfrac{4x^2+2x+x^3-1}{2x^5-2x^2}\)
\(=\dfrac{x^3+4x^2+2x-1}{2x^5-2x^2}\)
\(b,\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x+1\right).\left(x^2-x+1\right)}\)
\(=\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x^2-x+1\right)}\)
\(=\dfrac{x+1\left(x+1\right).\left(x^2-x+1\right)-\left(x^2+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1+x^3+1-x^2-2}{\left(x+1\right).\left(x^2-x+1\right)}\)
\(=\dfrac{x+0+x^3-x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x\left(1+x^2-x\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)
\(=\dfrac{x}{x+1}\)
Ta có: \(\left(x+z\right)\left(y+z\right)=1\)
\(\Rightarrow\left(x+z\right)^2\left(y+z\right)^2=1\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y+z\right)^2}+\dfrac{1}{\left(z+x\right)^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(y+z\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(z+x\right)^2}\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2+\left(y+z\right)^2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2-2\left(x+z\right)\left(y+z\right)+\left(y+z\right)^2+2\) (Vì: (x+z)(y+z)=1 =>2(x+z)(y+z)=2 )
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z-y-z\right)^2+2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\)
Áp dụng bất đẳng thức Cauchy, ta có :
\(\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2\ge2\sqrt{\dfrac{1}{\left(x-y\right)^2}\cdot\left(x-y\right)^2}=2\cdot1=2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\ge2+2=4\)
Vậy \(MinP=4\) khi \(x-y=1\); \(y+z=\dfrac{\sqrt{5}-1}{2}\); \(x+z=\dfrac{2}{\sqrt{5}-1}\)