Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

ta có : \(A=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{\dfrac{1}{36}}}=\dfrac{\sqrt{x}+\sqrt{y}}{\dfrac{1}{6}}=5\)

\(\Rightarrow\sqrt{x}+\sqrt{y}=5.\dfrac{1}{6}=\dfrac{5}{6}\)

\(\Rightarrow\sqrt{x};\sqrt{y}\) là nghiệm của phương trình \(X^2-\dfrac{5}{6}X+\dfrac{1}{36}=0\)

\(\Leftrightarrow36X^2-30X+1=0\)

\(\Delta'=\left(-15\right)^2-36.1=225-36=189>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(X_1=\dfrac{15-\sqrt{189}}{36}=\dfrac{5-\sqrt{21}}{12}\)

\(X_2=\dfrac{15+\sqrt{189}}{36}=\dfrac{5+\sqrt{21}}{12}\)

\(X=\sqrt{x}=\dfrac{5-\sqrt{21}}{12}\Rightarrow x;y=\left(\dfrac{5-\sqrt{21}}{12}\right)^2\)

\(X=\sqrt{x}=\dfrac{5-\sqrt{21}}{12}\Rightarrow x;y=\left(\dfrac{5+\sqrt{21}}{12}\right)^2\)

Ta có :

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

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

Đáp án 2+\(\sqrt{3}\) là đáp án cuối cùng đó bạn

Hint:\(Max_\text{(fx)}=5.. \text{tại}.. x=1\)

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+.....+\dfrac{2008}{2009!}=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+.....+\dfrac{2009-1}{2009!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+.....+\dfrac{2009}{2009!}-\dfrac{1}{2009!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+.......+\dfrac{1}{2008!}-\dfrac{1}{2009!}\)

\(=1-\dfrac{1}{2009!}\)

Ta chứng minh công thức:

\(1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}=\left(1+\dfrac{1}{n}+\dfrac{1}{n+1}\right)^2\) bằng cách quy đồng biểu thức ở vế phải rồi áp dụng vào bài tập

\(S=\dfrac{4}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{6}{2.3.4}-\dfrac{1}{2.3.4}+...+\dfrac{4018}{2008.2009.2010}-\dfrac{1}{2008.2009.2010}\)

\(=\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2008.2010}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\right)\)

\(=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)-\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2008.2009.2010}\right)\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)

\(=\left(1-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2009.2010}\right)\)

\(=1-\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2009.2010}\)

\(=\dfrac{1}{2010}\left(\dfrac{1}{2009}-1\right)-\left(\dfrac{1}{2009}-1\right)\)

\(=\left(\dfrac{1}{2010}-1\right)\left(\dfrac{1}{2009}-1\right)=\dfrac{2009}{2010}.\dfrac{2008}{2009}=\dfrac{1004}{1005}\)

\(VT=\dfrac{xy\sqrt{z-2}+yz\sqrt{x-3}+xz\sqrt{y-4}}{xyz}\)

\(=\dfrac{\sqrt{z-2}}{z}+\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{\sqrt{z-2}}{z}=\dfrac{\sqrt{2\left(z-2\right)}}{\sqrt{2}z}\le\dfrac{\dfrac{2+z-2}{2}}{\sqrt{2}z}=\dfrac{1}{2\sqrt{2}}\)

\(\dfrac{\sqrt{x-3}}{x}=\dfrac{\sqrt{3\left(x-3\right)}}{\sqrt{3}x}\le\dfrac{\dfrac{3+x-3}{2}}{\sqrt{3}x}=\dfrac{1}{2\sqrt{3}}\)

\(\dfrac{\sqrt{y-4}}{y}=\dfrac{\sqrt{4\left(y-4\right)}}{\sqrt{4}y}\le\dfrac{\dfrac{4+y-4}{2}}{\sqrt{4}y}=\dfrac{1}{2\sqrt{4}}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=\dfrac{\sqrt{z-2}}{z}+\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}\le\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{2\sqrt{4}}=VP\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=6\\y=8\\z=4\end{matrix}\right.\)

hay cái j cha nội ngược dấu kìa :v

\(a,\Leftrightarrow4x\le26244\\ \Leftrightarrow x\le6561\)

\(b,\Leftrightarrow4x\ge10\\ \Leftrightarrow x\ge2.5\)

\(c,\Leftrightarrow9x=12\\ \Leftrightarrow x=\dfrac{4}{3}\)

\(VT=x+2\sqrt{2x-4}\)

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

\(=\left(\sqrt{x-2}+\sqrt{2}\right)^2=VP\left(\text{đ}pcm\right)\)