Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{x^2-1}\)
= \(-\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\)
= \(\frac{-x-1+x-1+2}{\left(x-1\right)\left(x+1\right)}=0\)
c) \(\left(\frac{x^2-16}{x^2+8x+16}+\frac{6}{x+4}\right)\cdot\frac{2x}{x+2}\)
= \(\left(\frac{x^2-16}{\left(x+4\right)^2}+\frac{6\left(x+4\right)}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)
= \(\left(\frac{x^2-16+6x+24}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)
= \(\frac{x^2+6x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x-2}\)
= \(\frac{x^2+4x+2x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}\)
= \(\frac{\left(x+4\right)\left(x+2\right)}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}=\frac{2x}{x+4}\)
\(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\)
\(P=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\left(1\right)\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\frac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\frac{3}{2}\)
Từ (1) và (2)
\(\Rightarrow\frac{x^2}{xy+xz}+\frac{y^2}{xy+zy}+\frac{z^2}{xz+yz}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\)
\(\Leftrightarrow P\ge\frac{3}{2}\)
Vậy \(P_{min}=\frac{3}{2}\)
Dấu " = " xảy ra khi x = y= z
Áp dụng BĐT Netbitt ta có Vì x,y,z >0 nên
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\)
Dấu ''='' xảy ra khi x = y = z > 0
Théo bđt Cauchuy Schwarz dạng Engel ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{3}\)
Đk: x,y,z khác 0.
ta có: \(\left(y-z\right)^2\ge0\Rightarrow y^2+z^2\ge2yz\Leftrightarrow x^2+y^2+z^2\ge x^2+2yz\Leftrightarrow\frac{yz}{x^2+2yz}\ge\frac{yz}{x^2+y^2+z^2}\)
tương tự thì \(A\ge\frac{xy}{x^2+y^2+z^2}+\frac{yz}{x^2+y^2+z^2}+\frac{xz}{x^2+y^2+z^2}=\frac{xy+yz+xz}{x^2+y^2+z^2}\)
từ đề bài =>\(\frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
=> A =0
1/y+1/x+1/z=0
=>xy+yz+xz=0(tự cm)
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2=0
x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3xyz=3xyz
x^6+y^6+z^6=(x^2+y^2+z^2)(X^4+y^4+z^4+x^2y^2+y^2z^2+z^2z^2)+3(xyz)^2=3(xyz)^2
=> (x^6+y^6+z^6)/(x^3+y^3+z^3)=3(Xyz)^2/3xyz=xyz(dpcm)
:D???? ể??
\(x+y+z=0\Rightarrow\hept{\begin{cases}x=-y-z\\y=-z-x\\z=-x-y\end{cases}}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\)
\(\hept{\begin{cases}xy=\left(-y-z\right).y=-y^2-zy\\yz=\left(-x-z\right).z=-z^2-xz\\xz=\left(-y-x\right).x=-x^2-xy\end{cases}}\Rightarrow xy+yz+zx=-\left(x^2+y^2+z^2+xz+xy+zy\right)=0\)
\(\Leftrightarrow x=y=z=0??????\)
p/s: ko biết t lỗi hay đề lỗi ((:
Đặt a = y + z; b = z+ x; c = x+ y (a;b;c > 0)
=> x+ y + z = (a+b+c)/2
=> x= (a+b+c)/2 - a = (b+c- a)/2
y = (a+b+c)/2 - b = (a+c-b)/2; z = (a+b - c)/ 2
Khi đó \(P=\frac{b+c-a}{2a}+\frac{a+c-b}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}.\left(\frac{b}{a}+\frac{c}{a}-1+\frac{a}{b}+\frac{c}{b}-1+\frac{a}{c}+\frac{b}{c}-1\right)\)
=> \(P=\frac{b+c-a}{2a}+\frac{a+c-b}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}.\left(\left(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)-3\right)\right)\)
AD BĐT Cô - si có: \(\frac{a}{b}+\frac{b}{a}\ge2;\frac{b}{c}+\frac{c}{b}\ge2;\frac{c}{a}+\frac{a}{c}\ge2\)
=> \(P\ge\frac{1}{2}.\left(2+2+2-3\right)=\frac{3}{2}\)=> Min P = 3/2
Dấu "=" khi a = b = c<=> x = y = z
Áp dụng Cauchy - Schwarz và AM-GM :
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{xz+yz}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\)
\(\ge\frac{\left(x+y+z\right)^2}{\frac{2\left(x+y+z\right)^2}{3}}=\frac{3}{2}\)
Đẳng thức xảy ra tại x=y=z