Ta có:

\(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)

\(\Leftrightarrow\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}=-2\)

\(\Leftrightarrow x^2z+x^2y+y^2x+y^2z+z^2x+z^2y+2xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=-z\\z=-x\end{cases}}\)

Với \(x=-y\)

\(\Rightarrow x^3+y^3+z^3=1\)

\(\Rightarrow z=1\)

\(\Rightarrow P=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x}+\frac{1}{-x}+\frac{1}{1}=1\)

Tương tự cho các trường hợp còn lại.

ĐK \(x;y\ge0\)

Ta có hệ \(\hept{\begin{cases}\sqrt{x}+\sqrt{y}=10\left(1\right)\\\sqrt{x+6}+\sqrt{y+6}=14\left(2\right)\end{cases}}\)

Từ (1) ta có \(\sqrt{x}+\sqrt{y}=10\Rightarrow x+y+2\sqrt{xy}=100\Rightarrow x+y=100-2\sqrt{xy}\)

Từ (2) ta có \(\sqrt{x+6}+\sqrt{y+6}=14\Rightarrow x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}=196\)

\(\Rightarrow100-2\sqrt{xy}+12+2\sqrt{\left(x+6\right)\left(y+6\right)}=196\)

\(\Rightarrow\sqrt{\left(x+6\right)\left(y+6\right)}=42+\sqrt{xy}\Rightarrow xy+6x+6y+36=1764+84\sqrt{xy}+xy\)

\(\Leftrightarrow6\left(x+y\right)=1728+84\sqrt{xy}\Rightarrow6\left(100-2\sqrt{xy}\right)=1728+84\sqrt{xy}\)

\(\Leftrightarrow-1128=96\sqrt{xy}\left(ktm\right)\)

Vậy hệ vô nghiệm 

bài này khó quá à

\(\sqrt{x\left(x-2\right)}+\sqrt{x\left(x-5\right)}=\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}\right)=0\)

TH1: x = 0 (nhận)

TH2:

\(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-2\right)+\left(\sqrt{x-5}-1\right)-\left(\sqrt{x+3}-3\right)=0\)

\(\Leftrightarrow\frac{x-2-4}{\sqrt{x-2}+2}+\frac{x-5-1}{\sqrt{x-5}+1}-\frac{x+3-9}{\sqrt{x+3}+3}=0\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2}+2}+\frac{1}{\sqrt{x-5}+1}-\frac{1}{\sqrt{x+3}+3}\right)\left(x-6\right)=0\)

Pt \(\frac{1}{\sqrt{x-2}+2}+\frac{1}{\sqrt{x-5}+1}-\frac{1}{\sqrt{x+3}+3}=0\) vô n_o

=> x - 6 = 0

<=> x = 6 (nhận)

x^2+x+y^2+y+z^2+z<=18 suy ra (x+y+z)^2/3+x+y+z<=18

Đặt x+y+z=t thì t^2/3+t-18<=0 suy ra t^2+3t-54<=0>>>(t+9)(t-6)<=0>>>t-<=0>>>t<=6

P>=(1+1+1)^2/2x+2y+2z+3(BĐT Cauchuy-Swartch)=9/2(x+y+z)+3>=9/2.6+3=9/15=3/5

Dấu = khi x=y=z=2(tính dấu = của BĐT Cauchuy-Swartch nhé)

giống cách mình,mà đó là schwarts mà Hoàng Minh Hoàng

\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)

\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{2x}{\sqrt{x}-2}\)

\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)

\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)

\(=\frac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)\(=\frac{-\sqrt{x}}{\sqrt{x}+1}.\left(x-1\right)=\frac{-x\sqrt{x}+\sqrt{x}}{\sqrt{x}+1}\)

\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)

\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)

\(=\frac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)\(=\frac{-\sqrt{x}}{\sqrt{x}+1}.\left(x-1\right)=\frac{-x\sqrt{x}+\sqrt{x}}{\sqrt{x}+1}\)

Đặt \(\sqrt{x}=a;\sqrt{y}=b\left(a,b>0\right)\) thì có:

\(\hept{\begin{cases}a^3+b^3=2ab\\a+b=2\end{cases}}\). Khi đó xét pt(1)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=2\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2-2\right)=0\)

*)Xét \(a+b=0\Rightarrow a=-b\Rightarrow a=b=0\) (loại)

*)Xét \(a^2-ab+b^2-2=0\Rightarrow a^2+b^2-ab=2\)

Do \(a,b\ge0\) nên xài AM-GM ta có:

\(a^2+b^2\ge2ab\Rightarrow a^2+b^2-ab\ge ab=2\)

Và \(ab\le\frac{\left(a+b\right)^2}{4}=2\) Xảy ra khi \(a=b=1\) (thỏa)

Vậy nghiệm hpt là \(a=b=1\)

Đặt √x=a;√y=b,ta có;a^3+b^3=2ab;a+b=2>>>(a+b)(a^2-ab+b^2)=2(a^2-ab+b^2)=2ab

a^2-ab+b^2=ab >>>(a-b)^2=0 >>>a=b>>>x=y=1

Đk \(x;y\ge0\)

Ta có hệ \(\hept{\begin{cases}\sqrt{x}+\sqrt{y}=4\left(1\right)\\\sqrt{x+5}+\sqrt{y+5}=6\left(2\right)\end{cases}}\)

Từ (1) ta có \(\sqrt{x}+\sqrt{y}=4\Rightarrow x+y+2\sqrt{xy}=16\Rightarrow x+y=16-2\sqrt{xy}\)

Từ (2) ta có \(\sqrt{x+5}+\sqrt{y+5}=6\Rightarrow x+5+y+5+2\sqrt{\left(x+5\right)\left(y+5\right)}=36\)

\(\Rightarrow16-2\sqrt{xy}+10+2\sqrt{\left(x+5\right)\left(y+5\right)}=36\)

\(\Leftrightarrow2\sqrt{\left(x+5\right)\left(y+5\right)}=10+2\sqrt{xy}\)

\(\Leftrightarrow4\left(xy+5x+5y+25\right)=100+40\sqrt{xy}+4xy\)

\(\Leftrightarrow x+y-2\sqrt{xy}=0\Leftrightarrow\sqrt{x}=\sqrt{y}\Leftrightarrow x=y\)

\(\Rightarrow\sqrt{x}+\sqrt{x}=4\Leftrightarrow x=4\Rightarrow y=4\)

Vậy  hệ có nghiệm \(\left(x;y\right)=\left(4;4\right)\)

 bài này gần giống

Cho tam giác ABC , M nằm trong tam giác các dg thẳng AM,BM,CM lần lượt cắt các cạnh BC,AC,AB tại A1,B1,C1?

xác đinh vị trí của M để 
a, AM/A1M + BM/B1M +CM/ C1M đạt GTNN 
b, AM/A1M . BM/B1M . CM/ C1M đạt GTNN 
c, A1M/AM + B1M/BM +C1M/CM đạt GTLN 
d, A1M/AM . B1M/BM . C1M/CM đạt GTLN

Bài làm

Đặt S(MBC) =S1, S(AMC) =S2; S(AMB) =S3 

AM/A1M = S3/S(A1BM) = S2/S(CA1M) = (S2+S3)/S1 

Tương tự 
BM/B1M = (S1+S3)/S2 
CM/C1M = (S1+S2)/S3 

=> AM/A1M + BM/B1M +CM/C1M = (S2+S3)/S1 + (S1+S3)/S2 + (S1+S3)/S2 
=(S2/S1 + S1/S2) + (S3/S1+S1/S3) + (S2/S3+S3/S2) >= 6 

Khi M là trong tâm 

b] 

AM/A1M . BM/B1M . CM/ C1M= (S2+S3)/S1 + (S1+S3)/S2 + (S1+S3)/S2 >=8 (Cauchy) 
Khi M là trọng tâm 

c] 

A1M/AM + B1M/BM +C1M/CM = S1/(S2+S3) +S2/(S1+S3) + S3/(S2+S1)= 
= (S1+S2+S3) [1/(S2+S3) +1/(S1+S3) + 13/(S2+S1)] -3= 
=1/2[(S1+S2) + (S2+S3) +(S3+S1)][1/(S2+S3) +1/(S1+S3) + 1/(S2+S1)] -3>= 
>= 9/2 -3 =3/2 

Khi M là trong tâm 


d] Hệ quả từ B Max =1/8