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Vào lúc: 2019-11-15 21:08:32 Xem câu hỏi

Ta có:

\(A=a^2+b^2+c^2=\left(a-b\right)a+\left(b-c\right)\left(a+b\right)+c\left(a+b+c\right)\) 

\(\Rightarrow A\ge5\left(a-b\right)+6\left(b-c\right)+7c=5a+b+c=4a+\left(a+b+c\right)\ge27\)

Dau '=' xay ra khi \(a=5,b=1,c=1\)

Vào lúc: 2019-11-15 12:53:57 Xem câu hỏi

\(ĐK:1-x-2x^2\ge0\)

Ta có:

Min

\(B=\frac{x}{2}+\sqrt{1-x-2x^2}=\left(\sqrt{x+1}+\frac{\sqrt{1-2x}}{2}\right)^2-\frac{5}{4}\text{ }\ge-\frac{5}{4}\)

Dau '='' xay ra khi \(x=-\frac{1}{2}\)

Max

Ta có:

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

Da '=' xay ra khi \(x=0\)

Vào lúc: 2019-11-14 13:03:53 Xem câu hỏi

1.

\(ĐK:x\ne0\)

HPT

\(\Leftrightarrow\hept{\begin{cases}2x\left(x+y\right)-3x+1=0\\3x\left(x+y\right)-x-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x\left(x+y\right)-\frac{9}{2}x+\frac{3}{2}=0\left(1\right)\\3x\left(x+y\right)-x-2=0\left(2\right)\end{cases}}\)

\(\left(1\right)-\left(2\right)\Leftrightarrow\frac{7}{2}x=\frac{7}{2}\)

\(\Leftrightarrow x=1\left(3\right)\)

\(\left(1\right),\left(3\right)\Rightarrow3\left(1+y\right)-3=0\)

\(\Leftrightarrow y=0\)

Vay nghiem cua HPT la \(\left(1;0\right)\)

Vào lúc: 2019-11-14 12:54:38 Xem câu hỏi

different way

Áp dụng min-cop-xki ta có:

\(M=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}\ge\sqrt{\left(x+y\right)^2+\left(\frac{1}{x}+\frac{1}{y}\right)^2}\ge\sqrt{16+\frac{16}{\left(x+y\right)^2}}=\sqrt{17}\)

Dau '=' xay ra khi \(x=y=2\)

Vào lúc: 2019-10-27 19:13:02 Xem câu hỏi

Ta di chung minh

\(\frac{1}{x^2+y^2+1}+\frac{1}{y^2+z^2+1}+\frac{1}{z^2+x^2+1}\le1\)

\(\Leftrightarrow\frac{x^2+y^2}{x^2+y^2+1}+\frac{y^2+z^2}{y^2+z^2+1}+\frac{z^2+x^2}{z^2+x^2+1}\ge2\)

\(VT\ge\frac{\left(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\right)^2}{2\left(x^2+y^2+z^2\right)+3}\left(1\right)\)

Gio chung minh:

\(VT_{\left(1\right)}\ge2\)

\(\Leftrightarrow\left(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\right)^2\ge4\left(x^2+y^2+z^2\right)+6\)

\(\Leftrightarrow\sqrt{\left(x^2+y^2\right)\left(y^2+z^2\right)}+\sqrt{\left(y^2+z^2\right)\left(z^2+x^2\right)}+\sqrt{\left(z^2+x^2\right)\left(x^2+y^2\right)}\ge x^2+y^2+z^2+3\left(2\right)\)

Ta co:

\(\sqrt{\left(x^2+y^2\right)\left(y^2+z^2\right)}=\sqrt{\left(x^2+y^2\right)\left(z^2+y^2\right)}\ge zx+y^2\)

The same

\(\Rightarrow VT_2\ge x^2+y^2+z^2+xy+yz+zx\)

Chung minh:

\(VT_2\ge x^2+y^2+z^2+3\)

\(\Leftrightarrow xy+yz+zx\ge3\)

Ta lai co:

\(xy+yz+zx\ge3\sqrt[3]{\left(xyz\right)^2}=3\)

Dau '=' xay ra khi \(x=y=z=1\)

Vào lúc: 2019-10-23 18:18:21 Xem câu hỏi

PT

\(\Leftrightarrow20y^2-150=3x\left(2y-5\right)\)

\(\Leftrightarrow3x=\frac{20y^2-150}{2y-5}\)

De \(x\in Z\Rightarrow\frac{20y^2-150}{2y-5}\in Z\)

Dat \(M=\frac{20y^2-150}{2y-5}=5\left(2y+5\right)-\frac{25}{2y-5}\)

De \(3x=M=10y+25-\frac{25}{2y-5}\in Z\Rightarrow\frac{25}{2y-5}\in Z\Rightarrow2y-5\in\left\{-5;-1;1;5\right\}\)

Ta tim duoc

\(y_1=0;y_2=2;y_3=3;y_4=5\)

\(\Rightarrow x_1=x_3=30;x_2=70;x_4=70\)

Vào lúc: 2019-10-22 17:26:22 Xem câu hỏi

Ta co:

\(Q=a^3+b^3+c^3=\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)-6\ge3\left(a+b+c\right)-6=3\)

Dau '=' xay ra khi \(a=b=c=1\)

Vay \(Q_{min}=3\)khi \(a=b=c=1\)

Vào lúc: 2019-10-21 11:34:52 Xem câu hỏi

\(DK:x\in\left[\frac{1}{2};4\right]\)

PT

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

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

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

Vì \(\frac{x+2}{\sqrt{x^2+x+2}}+\frac{2}{\sqrt{2x-1}+1}+1>0\)

\(\Rightarrow x=1\left(n\right)\)

Vay nghiem cua PT la \(x=1\)

Vào lúc: 2019-10-19 20:20:44 Xem câu hỏi

PT

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(x+5\right)=m\)

\(\Leftrightarrow\left(x^2+4x+3\right)\left(x^2+4x-5\right)=m\)

\(\Leftrightarrow\left(x^2+4x-1+4\right)\left(x^2+4x-1-4\right)=m\)

\(\Leftrightarrow\left(x^2+4x-1\right)^2-16=m\)

\(\Leftrightarrow\left(x^2+4x-1\right)^2=m+16\) \(\left(DK:m\ge-16\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+4x-1=\sqrt{m+16}\left(1\right)\\x^2+4x-1=-\sqrt{m+16}\left(2\right)\end{cases}}\)

PT(1)

\(\Leftrightarrow x^2+4x-1-\sqrt{m+16}=0\)

Ta co:

\(\Delta^`=2^2-1.\left(-1-\sqrt{m+16}\right)=5+\sqrt{m+16}>0\)

\(\Rightarrow\hept{\begin{cases}x_1=-2+\sqrt{5+\sqrt{m+16}}\\x_2=-2-\sqrt{5+\sqrt{m+16}}\end{cases}}\)

PT(2)

\(\Leftrightarrow x^2+4x-1+\sqrt{m+16}=0\)

Ta lai co:

\(\Delta^`=2^2-1.\left(-1+\sqrt{m+16}\right)=5-\sqrt{m+16}\)

De PT co 4 nghiem phan biet thi PT(1) va PT(2) co 2 nghiem phan bet

Suy ra PT(2) co 2 nghiem phan biet khi 

\(5-\sqrt{m+16}>0\)

\(\Leftrightarrow m< 9\)

\(\Rightarrow\hept{\begin{cases}x_3=-2+\sqrt{5-\sqrt{m+16}}\\x_4=-2-\sqrt{5-\sqrt{m+16}}\end{cases}}\)

Ta lai co:

\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_4}+\frac{1}{x_5}=\frac{x_1+x_2}{x_1x_2}+\frac{x_4+x_5}{x_4x_5}=\frac{4}{1+\sqrt{m+16}}+\frac{4}{1-\sqrt{m+16}}\text{ }=-\frac{8}{15+m}\)\(\left(DK:m\ne-15\right)\)

Ma \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=-1\)

\(\Leftrightarrow-\frac{8}{m+15}=-1\)

\(\Leftrightarrow m=-7\)

Vay de PT \(\left(x^2-1\right)\left(x+3\right)\left(x+5\right)=m\)co 4 gnhiem phan biet thoa man 

\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=-1\)thi m=-7

Vào lúc: 2019-10-19 19:36:30 Xem câu hỏi

PT

\(log2^{x^2-x}+log3^{x^2-x}=log2.5^{x^2-x}\)

\(\Leftrightarrow x^2-xlog2+x^2-xlog3=2\left(x^2-x\right)log5\)

\(\Leftrightarrow\left(x^2-x\right)\left(log2+log3-2log5\right)=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vay nghiem cua PT la \(x=0\)va \(x=1\)

Vào lúc: 2019-10-16 17:58:56 Xem câu hỏi

2.

\(DK:\hept{\begin{cases}x\ge-\frac{1}{5}\\x\ne0\end{cases}}\)

PT

\(\Leftrightarrow6+3\sqrt{5x+1}\left(\sqrt{5x+1}-1\right)=14\left(\sqrt{5x+1}-1\right)\)

\(\Leftrightarrow15x+23-17\sqrt{5x+1}=0\)

\(\Leftrightarrow\left(68-17\sqrt{5x+1}\right)+\left(15x-45\right)=0\)

\(\Leftrightarrow\frac{17\left(x-3\right)}{4+\sqrt{5x+1}}+15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{17}{4+\sqrt{5x+1}}+15\right)=0\)

Vi \(\frac{17}{4+\sqrt{5x+1}}+15>0\)

\(\Rightarrow x=3\left(n\right)\)

Vay nghiem cua PT la \(x=3\)

Vào lúc: 2019-10-16 17:46:06 Xem câu hỏi

\(DK:x\ge\frac{1}{2}\)

PT

\(\Leftrightarrow\left(4x-4\right)+\left(\sqrt{2x-1}-1\right)=0\)

\(\Leftrightarrow4\left(x-1\right)+\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4+\frac{2}{\sqrt{2x-1}+1}\right)=0\)

Vi \(4+\frac{2}{\sqrt{2x-1}+1}>0\)

\(\Rightarrow x=1\left(n\right)\)

Vay nghiem cua PT la \(x=1\)

Vào lúc: 2019-10-16 12:05:53 Xem câu hỏi

1.

\(DK:x\ge2\)

PT

\(\Leftrightarrow\left(2+x\right)\sqrt{x-2}-\left(x+2\right)\left(x-2\right)\)

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

Cho này thì ok ròi nhé

2.

\(DK:x\le\frac{5}{2}\)

Xet \(x\in\left[0;\frac{5}{2}\right]\)

PT

\(\Leftrightarrow x^2-4x=5-2x\)

\(\Leftrightarrow x^2-2x-5=0\)

Ta co:

\(\Delta^`=\left(-1\right)^2-1.\left(-5\right)=6>0\)

\(\Rightarrow\hept{\begin{cases}x_1=1+\sqrt{6}\left(l\right)\\x_2=1-\sqrt{6}\left(l\right)\end{cases}}\)

Xet \(x\le0\)

PT

\(4x-x^2=5-2x\)

\(\Leftrightarrow x^2-6x+5=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=5\left(l\right)\end{cases}}\)

Vay PT vo nghiem 

Vào lúc: 2019-10-15 18:49:38 Xem câu hỏi

\(DK:x\notin\left(\frac{1-\sqrt{5}}{2};\frac{1+\sqrt{5}}{2}\right)\)

PT

\(\Leftrightarrow\left(\sqrt{8x+1}-5\right)-\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\frac{8\left(x-3\right)}{\sqrt{8x+1}+5}-\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{8}{\sqrt{8x+1}}-x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(1\right)\\\frac{8}{\sqrt{8x+1}+5}-x-2=0\left(2\right)\end{cases}}\)

\(\Leftrightarrow\left(x+2\right)\left(\sqrt{8x+1}+5\right)=8\left(DK:x>-2\right)\)

\(\Leftrightarrow\left(x+2\right)\sqrt{8x+1}+x+2=0\)

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{8x+1}+\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\sqrt{8x+1}+\sqrt{x+2}=0\)

\(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{8}\\x=-2\end{cases}}\left(KTM\right)\)

Vay nghiem cua PT la \(x=3\)

Vào lúc: 2019-10-14 11:53:00 Xem câu hỏi

\(DK:\hept{\begin{cases}-1\le x\le1\\x\ne0\end{cases}}\)

Ta co:

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

Suy ra: f(x) la ham so chan

Vào lúc: 2019-10-14 11:44:19 Xem câu hỏi

HPT\(\Leftrightarrow\hept{\begin{cases}x=3m+2-y\\3\left(3m+2-y\right)-2y+m-11=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3m+2-y\\-5y+10m-5=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3m+2-\left(2m-1\right)\\y=2m-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=m+3\\y=2m-1\end{cases}}\)

Ta co:

\(x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)

Dau '=' xay ra khi \(m=\frac{5}{3}\)

\(\Rightarrow\left(x;y\right)=\left(\frac{14}{3};\frac{7}{3}\right)\)

Vay cap nghiem (x;y) de \(x^2-y^2\)dat max la \(\left(\frac{14}{3};\frac{7}{3}\right)\)

Vào lúc: 2019-10-14 11:27:04 Xem câu hỏi

Ta co:

\(abc\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)

\(=\frac{2a\left(1+a^2\right)2b\left(1+b^2\right)2c\left(1+c^2\right)}{8}\le\frac{\frac{\left[\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\right]^2}{64}}{8}\le\frac{\frac{\left(a+b+c+3\right)^{12}}{27^4}}{512}=\frac{\frac{6^{12}}{27^4}}{512}=8\)

Dau '=' xay ra khi \(a=b=c=1\)

Vào lúc: 2019-10-13 07:39:09 Xem câu hỏi

Ta co:

\(x+y\ge x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

\(\Rightarrow\frac{\left(x+y\right)^2}{2}\le x+y\)

\(\Leftrightarrow x+y\le2\)

Dau '=' xay ra khi \(x=y=1\)

Vào lúc: 2019-10-13 07:28:04 Xem câu hỏi

Ta co:

\(\left(1+a^2\right)^2\le\left(1+a\right)\left(1+a\right)=\left(1+a\right)^2\)

\(\Rightarrow1+a^2\le1+a\)

The same:

\(1+b^2\le1+b\)

\(1+c^2\le1+c\)

\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\le\left(1+a\right)\left(1+b\right)\left(1+c\right)\)

\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\le\frac{\left(3+a+b+c\right)^3}{27}=\frac{6^3}{27}=8\)

Ta lai co:

\(abc\le\frac{\left(a+b+c\right)^3}{27}=\frac{27}{27}=1\)

\(abc\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\le8\)

Dau '=' xay ra khi \(a=b=c=1\)

Vào lúc: 2019-10-12 18:36:22 Xem câu hỏi

PT\(\Leftrightarrow\left(y^4-x^2\right)+\left(y^2+x\right)=-4\)

\(\Leftrightarrow\left(y^2+x\right)\left(y^2-x+1\right)=-4\)

Ma \(-4=\left(-4\right).1=1.\left(-4\right)=\left(-2\right).2=2.\left(-2\right)\)

TH1:

\(\hept{\begin{cases}y^2+x=-4\\y^2-x+1=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-4-y^2\\y^2=x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-2\\y^2=x\end{cases}}\)(loai)

TH2:

\(\hept{\begin{cases}y^2+x=1\\y^2-x+1=-4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1-y^2\\y^2=-2\end{cases}}\)(loai)

TH3:

\(\hept{\begin{cases}y^2+x=-2\\y^2-x+1=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y^2+x=-2\\y^2=-\frac{1}{2}\end{cases}}\)(loai)

TH4;

\(\hept{\begin{cases}y^2+x=2\\y^2-x+1=-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y^2+x=2\\y^2=-\frac{1}{2}\end{cases}}\)(loai)

Vay khong co nghiem nguyen nao thoa man PT

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