cho x,y,z \(\in\)R*
có: \(x^3+y^3+z^3=1\)và \(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)
tính P=\(\frac{1}{x}+\frac{1}{z}+\frac{1}{y}\)
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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
\(\sqrt{x\left(x-2\right)}+\sqrt{x\left(x-5\right)}=\sqrt{x\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}\right)=0\)
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 no
=> 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
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.