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\(y+z=-x\)
\(\left(y+z\right)^5=-x^5\)
\(y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(x^5+y^5+z^5+5yz\left[\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right]=0\)
\(x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)-5xyz\left(\left(y^2+2yz+z^2\right)+y^2+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
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Ta có: x + y + z = 0 <=> y + z = -x
(y+z)5 = (-x)5
y5 + z5 + 5y4z + 10y3z2 + 10y2z3 + 5yz4 = -x5
y5 + z5 + 5y4z + 10y3z2 + 10y2z3 + 5yz4 + x5 = 0
x5 + y5 + z5 +5xyz[ y3 + 2y2z + 2yz2 + z3 ] = 0
x5 + y5 + z5 + 5xyz[(y+z)(y2 -yz -z2)+ 2yz(x+z)] = 0
x5 + y5 + z5 +5xyz[(y+z)(y2 +yz + z2)] = 0
2.(x5 + y5 + z5) + 5xyz(y+z)(y2+yz+z2) - (x5 + y5 + z5) = 0
2(x5 + y5 + z5) - 5xyz[(y2+2yz+z2)+y2+z2] = 0
2(x5 + y5 + z5) = 5xyz[(y+z)2 + y2 + z2]
2(x5 + y5 + z5) = 5xyz[(-x)2 + y2 + z2]
2(x5 + y5 + z5) = 5xyz(x2 + y2 + z2).
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a, Ta có: \(2\left(x^8+y^8\right)\ge\left(x^3+y^3\right)\left(x^5+y^5\right)\)
\(\Leftrightarrow x^8+y^8\ge x^5y^3+x^3y^5\)
Ta CM: \(\Leftrightarrow x^8+y^8\ge x^5y^3+x^3y^5\)
Áp dụng bđt Cô si:
\(x^8+x^8+x^8+x^8+x^8+y^8+y^8+y^8\ge8x^5y^3\) (*)
Tương tự, \(5y^3+3x^3\ge8x^3y^5\) (**)
Từ (*), (**) \(\Rightarrowđpcm\)
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\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
Lấy (2) cộng (3) ta được
\(x^2+y^2-yz-zx=2\) (4)
Lấy (1) - (4) ta được
\(2x\left(x+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-z\end{matrix}\right.\)
Xét 2 TH rồi thay vào tìm được y và z
1. \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{6}{5}\\\dfrac{y+z}{yz}=\dfrac{3}{2}\\\dfrac{z+x}{zx}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{5}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{7}{10}\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
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A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
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a/ Đơn giản là dùng phép thế:
\(x+2y+x+y+z=0\Rightarrow x+2y=0\Rightarrow x=-2y\)
\(x+y+z=0\Rightarrow z=-\left(x+y\right)=-\left(-2y+y\right)=y\)
Thế vào pt cuối:
\(\left(1-2y\right)^2+\left(y+2\right)^2+\left(y+3\right)^2=26\)
Vậy là xong
b/ Sử dụng hệ số bất định:
\(\left\{{}\begin{matrix}a\left(\frac{x}{3}+\frac{y}{12}-\frac{z}{4}\right)=a\\b\left(\frac{x}{10}+\frac{y}{5}+\frac{z}{3}\right)=b\end{matrix}\right.\)
\(\Rightarrow\left(\frac{a}{3}+\frac{b}{10}\right)x+\left(\frac{a}{12}+\frac{b}{5}\right)y+\left(\frac{-a}{4}+\frac{b}{3}\right)z=a+b\) (1)
Ta cần a;b sao cho \(\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}=-\frac{a}{4}+\frac{b}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}\\\frac{a}{3}+\frac{b}{10}=-\frac{a}{4}+\frac{b}{3}\end{matrix}\right.\) \(\Rightarrow\frac{a}{2}=\frac{b}{5}\)
Chọn \(\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\) thay vào (1):
\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)
vì \(x+y+z=0\) \(\Rightarrow\)\(x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=-z^3\)\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=-3xy.\left(x+y\right)\)
\(\Rightarrow x^3+y^3+z^3=3xyz\) do \(x+y=-z\)
\(\Rightarrow\left(x^3+y^3+z^3\right).\left(x^2+y^2+z^2\right)=3xyz.\left(x^2+y^2+z^2\right)\)
\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(y^2+z^2\right)+y^3.\left(x^2+z^2\right)+z^3.\left(x^2+y^2\right)\)
lại có: \(x^2+y^2=\left(x+y\right)^2-2xy=z^2-2xy\)
tương tự thì: \(y^2+z^2=x^2-2yz\)
\(z^2+x^2=y^2-2xz\)
vì vậy nên \(3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(x^2-2yz\right)+y^3.\left(y^2-2xz\right)+z^3.\left(z^2-2xy\right)\)
\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=2x^5+2y^5+2z^5-2xyz.\left(x^2+y^2+z^2\right)\)
\(\Rightarrow5xyz.\left(x^2+y^2+z^2\right)=2.\left(x^5+y^5+z^5\right)\)
đpcm
VÌ \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^5=-z^5\)
\(\Leftrightarrow x^5+y^{^5}+5\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right)=-z^5\)
\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x^3+y^3+2x^3y^2+2x^2y^3\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x+y\right)+\left(x^2-xy+y^2+2xy\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5-5xyz\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5=5xyz\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(2x^2+2xy+2y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+\left(x+y\right)^2+y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)Vì (x+y=-z)
HT