\(\hept{\begin{cases}x-y=1\\xy=6\end{cases}}\Leftrightarrow\hept{\begin{cases}y=x-1\\x\left(x-1\right)=6\end{cases}}\Leftrightarrow\hept{\begin{cases}y=x-1\\x^2-x-6=0\left(\cdot\right)\end{cases}}\)

Giải \(\left(\cdot\right)\), ta có \(x^2-x-6=0\)\(\Leftrightarrow x^2-3x+2x-6=0\)\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Nếu \(x=3\)thì \(y=x-1=3-1=2\)

Nếu \(x=-2\)thì \(y=-2-1=-3\)

Vậy hệ phương trình đã cho có 2 nghiệm \(\left(3;2\right)\)và \(\left(-2;-3\right)\)

Gọi 3 độ dài kích thước hình hộp chữ nhật là a;b;h .

Gọi độ dài 1 cạnh hình lập phương là c 

=> V_hhcn = a.b.h

V_hlp = c³ ; mà a + b + h = c + c + c = 3c

Khi đó V_hlp = c³ = \(\left(\frac{a+b+h}{3}\right)^3\ge\left(\frac{3\sqrt[3]{abh}}{3}\right)^3=abh\)= V_hhcn 

=> ĐPCM ("=" khi a = b = h = c)

a) Ta có \(V_{hhcn}=V_{hlp}\)

=> a.b.h = c³ 

Lại có : a + b + h \(\ge3\sqrt[3]{abh}=3\sqrt[3]{c^3}=3c\)

=> a + b + h \(\ge3c\)

=> ĐPCM 

\(2=\frac{2}{1}\Leftrightarrow\frac{2}{1}=\frac{x}{2}=\frac{6}{y}=\frac{z}{5}=\frac{8}{t}\)

*Xét \(\frac{2}{1}=\frac{x}{2}\)

\(\frac{2}{1}=\frac{x}{2}\Leftrightarrow\frac{2\cdot2}{1\cdot2}=\frac{x}{2}\Leftrightarrow\frac{4}{2}=\frac{x}{2}\Rightarrow x=4\)

*Xét\(\frac{2}{1}=\frac{6}{y}\)

\(\frac{2}{1}=\frac{6}{y}\Leftrightarrow\frac{2\cdot3}{1\cdot3}=\frac{6}{y}\Leftrightarrow\frac{6}{3}=\frac{6}{y}\Rightarrow y=3\)

*Xét \(\frac{2}{1}=\frac{z}{5}\)

\(\frac{2}{1}=\frac{z}{5}\Leftrightarrow\frac{2\cdot5}{1\cdot5}=\frac{z}{5}\Leftrightarrow\frac{10}{5}=\frac{z}{5}\Rightarrow z=10\)

*Xét \(\frac{2}{1}=\frac{8}{t}\)

\(\frac{2}{1}=\frac{8}{t}\Leftrightarrow\frac{2\cdot4}{1\cdot4}=\frac{8}{t}\Leftrightarrow\frac{8}{4}=\frac{8}{t}\Rightarrow t=4\)

Vậy giá trị \(x,y,z,t\)thỏa mãn là\({\begin{cases}x=4\\y=3\\z=10\\t=4\end{cases}}\)

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Answer:

a. \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)   ĐK: \(x\ge0;x\ne1\)

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

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

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

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

\(=\sqrt{x}\left(1-\sqrt{x}\right)\)

b. Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\1-\sqrt{x}>0\end{cases}}\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Do vậy \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)

c. \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)

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

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

\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)

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Đặt \(\sqrt{x-1}=a;\sqrt{x+1}=b\) \(\left(a;b\ge0;x\ge1\right)\)

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

<=> ab = a + b - x + 4

<=> 2ab = 2(a + b) - 2x + 8

<=> 2ab = 2(a + b) - a² - b² + 8

<=> (a + b)² - 2(a + b) + 1 = 9

<=> (a + b - 1)² = 9

<=> \(\orbr{\begin{cases}a+b=4\\a+b=-2\end{cases}}\Leftrightarrow a+b=4\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{x-1}=4\)

\(\Leftrightarrow\sqrt{x+1}=4-\sqrt{x-1}\)

\(\Leftrightarrow\hept{\begin{cases}x+1=x-1-8\sqrt{x-1}+16\\1\le x\le17\end{cases}}\Leftrightarrow\hept{\begin{cases}4\sqrt{x-1}=7\\1\le x\le17\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}16\left(x-1\right)=49\\1\le x\le17\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{65}{16}\\1\le x\le17\end{cases}}\Leftrightarrow x=\frac{65}{16}\left(tm\right)\)

ĐK : \(x>2009;y>2010;z>2011\)

PT <=> \(\frac{-4\sqrt{x-2009}+4}{x-2009}+\frac{-4\sqrt{y-2010}+4}{y-2010}+\frac{-4\sqrt{z-2011}+4}{z-2011}=-3\)

<=> \(\frac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}+\frac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}+\frac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}=0\)

<=> \(\hept{\begin{cases}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)

Vậy phương trình có 1 nghiệm duy nhất (x;y;z) = (2013 ; 2014 ; 2015)