x+y-1/z=y+z-1=z+x+2/y với x,y,z=0. Tìm x,y,z
help
Cho x,y,z>0 và x+y+z=1 . Tìm MinP = ∑ \(\dfrac{1}{x+y+1}\)
Cho x,y,z>0 và x+y+z =1 . Tìm Min A = ∑ \(\dfrac{x}{y^2+x^2+1}\)
\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Cho x,y,z>0 thỏa mãn 1/x+1/y+1/z=2015. Tìm GTLN của (x+y)/(x^2+y^2) + (y+z)/(y^2+z^2) + (z+x)/(z^2+x^2)
với x,y,z>0 CMR
(x^2+y^2+z^2)(1/x^2+1/y^2+1/z^2) >= (x+y+z)(1/x+1/y+1/z)
Với a;b;c dương ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
Lại có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)
Áp dụng:
\(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge\dfrac{1}{3}\left(x+y+z\right)^2.\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\)
\(=\dfrac{1}{9}\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(=\dfrac{1}{9}.9.\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dấu "=" xảy ra khi \(x=y=z\)
tìm x,y,z biết: x/ z+y+1=y/ x+z+1=z/ x+y+2=x+y+z{x,y,z khác 0}
tìm x,y,z biết : x/(z+y+1) = y/(x+z+1) = z-(x+y-z) = x+y+z ( x,y,z khác 0 )
Tìm x, y, z
x3/8= y3/27 = z3/64 và x2+2y2-3z2 = -650
x/ y+z+1 = y/ x+z+2 = z/y+z+3= x+y+z cho x, y, z <>0 thoả y+z+x/x= z+y-y/y= x+y-z/z
Tính B= (1+ x/y). (1+y/z) .(1+z/x)
Tìm x,y,z biết: \(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)(x,y,z khác 0)
Dùng tính chất tỉ lệ thức:
x+y+z = 0\(\frac{x}{\left(y+z+1\right)}=\frac{y}{\left(x+z+1\right)}=\frac{z}{\left(x+y-2\right)}=0\Rightarrow x=y=z=0\)
Áp dụng tính chất tỉ lệ thức:
\(x+y+z=\frac{x}{\left(y+z+1\right)}=\frac{y}{\left(x+z+1\right)}=\frac{z}{\left(x+y-2\right)}=\left(\frac{x+y+z}{2x+2y+2z}\right)=\frac{1}{2}\)
=> x+y+z = \(\frac{1}{2}\)
+) \(2x=y+z+1=\frac{1}{2}-x+1\Rightarrow x=\frac{1}{2}\)
+) \(2y=x+z+1=\frac{1}{2}-y+1\Rightarrow y=\frac{1}{2}\)
+) \(z=\frac{1}{2}-\left(x+y\right)=\frac{1}{2}-1=\frac{-1}{2}\)
TA CÓ: \(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{z+y+1+x+z+1+x+y-2}=\frac{1.\left(x+y+z\right)}{\left(1+1-2\right)+2x+2y+2z}\)
\(=\frac{1.\left(x+y+z\right)}{0+2.\left(x+y+z\right)}=\frac{1.\left(x+y+z\right)}{2.\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow\frac{x}{z+y+1}=\frac{1}{2}\)\(\Rightarrow2x=z+y+1\)\(\Rightarrow3x=x+z+y+1\)\(\Rightarrow3x=\frac{1}{2}+1\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
\(\frac{y}{x+z+1}=\frac{1}{2}\)\(\Rightarrow2y=x+z+1\Rightarrow3y=y+x+z+1\Rightarrow3y=\frac{1}{2}+1=\frac{3}{2}\Rightarrow y=\frac{1}{2}\)
\(\frac{z}{x+y-2}=\frac{1}{2}\)\(\Rightarrow2z=x+y-2\Rightarrow3z=x+y+z-2\Rightarrow3z=\frac{1}{2}-2=\frac{-3}{2}\Rightarrow z=\frac{-1}{2}\)
VẬY X= 1/2; Y= 1/2 ; Z= -1/2
CHÚC BN HỌC TỐT!!!!!!
Cho 1/x+y +1/y+z +1/z+x=0 Tính P=(y+z)(z+x)/(x+y)^2 + (x+y)(z+x)/(y+z)^2+ (y+z)(x+y)/(z+x)^2
Đặt \(\dfrac{1}{a}=\dfrac{1}{x+y},\dfrac{1}{b}=\dfrac{1}{y+z},\dfrac{1}{c}=\dfrac{1}{z+x}\)
Đề trở thành: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\), tính \(P=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tương đương \(ab+bc=-ac\)
\(P=\dfrac{b^3c^3+a^3c^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(ab+bc\right)\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}=\dfrac{-ac\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}\)
\(=\dfrac{a^2c^2-a^2b^2+ab^2c-b^2c^2}{ab^2c}=\dfrac{ac}{b^2}-\dfrac{a}{c}+1-\dfrac{c}{a}\)\(=ac\left(\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\right)-\dfrac{a}{c}+1-\dfrac{c}{a}\) (do \(\dfrac{1}{b}=-\dfrac{1}{a}-\dfrac{1}{c}\) tương đương \(\dfrac{1}{b^2}=\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\))
\(=3\)
Vậy P=3
1,Tìm GTLN: 1, A=|x^2-x+1|-|x^2-x-2|
2,Tìm GTLN: B=|x-y|+|x-z|+|y-z| với 0<x,y,z<3
\(Taco:\)
\(|x^2-x+1|-|x^2-x-2|=|x^2-x+1|+\left(-|x^2-x-2|\right)\)
\(\ge|x^2-x+1-x^2+x+2|=3\)
Dấu "=" xảy ra khi: \(\left(x^2-x+1\right)\left(x^2-x-2\right)\ge0\Leftrightarrow........\)