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D = \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\) . \(\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) = \(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(A=\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) \(:\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\) \(:\left[\frac{\sqrt{x}+2}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+x-7-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
Bổ sung giả thuyết x ,y \(\ge0\)
Do giả thiết x ,y \(\ge0\)
\(\sqrt{x}+\sqrt{y}\) =1
nên:
xy (x+y )\(^2\)\(\le\) \(\dfrac{1}{64}\)
<=> 64 xy (x + y )\(^2\) \(\le\)1
<=> 64 xy ( x + y)\(^2\)\(\le\)(\(\sqrt{x}+\sqrt{y}\))\(^8\)
<=> 64 xy ( x + y )\(^2\) < \((x+2\sqrt{xy}+y)^4\)
Áp dụng bất đẳng thức Cauchy cho 2 số không âm x + y và \(2\sqrt{xy}\)
ta có ;
x + y + 2\(\sqrt{xy}\) \(\ge\) \(2\sqrt{x+y}2\sqrt{xy}\)
=> ( x + y +2\(\sqrt{xy}\)) \(^4\)\(\ge\) (\(2\sqrt{x+y}2\sqrt{xy}\) )\(^4\)= 64 xy (x + y)\(^2\)
=> ĐIỀU PHẢI CHỨNG MINH
Dấu bằng xảy ra <=> x + y = \(2\sqrt{xy}\)
<=> x = y = \(\dfrac{1}{4}\)
ĐKXĐ: \(-1\le x\le1\)
Đặt \(a=\sqrt{1-x}>0\)
\(b=\sqrt{1+x}>0\)
\(\Rightarrow a^2+b^2=2\) và \(a^2-b^2=-2x\)
Khi đó: \(B=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{1-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)
\(=\frac{\sqrt{1-ab}\left(a+b\right)\left(2-ab\right)}{2-ab}\)\(\Rightarrow B=\sqrt{1-ab}\left(a+b\right)\Rightarrow B\sqrt{2}=\sqrt{2-2ab}\left(a+b\right)\)\(=\sqrt{a^2+b^2-2ab}\left(a+b\right)=\left(a-b\right)\left(a+b\right)=a^2-b^2=\)\(-2x\)
\(\Rightarrow b=-\sqrt{2}x\)
- Các ĐKXĐ tự tìm dùm mình hen :)
Ta có : \(D=\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}\)
=> \(D=\left(\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{1}{\sqrt{x}+2}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{5+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{1}{\sqrt{x}-3}\right)\left(\sqrt{x}-3\right)=1\)
Ta có : \(E=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{a-2\sqrt{a}+1}\)
=> \(E=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{\left(\sqrt{a}-1\right)^2}\)
=> \(E=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
=> \(E=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
( làm đến đây thôi câu còn lại bạn tự làm hen )
Ghét nhất mấy câu viết sai đề b, c sai rất nhiều bạn ới
đấy là mình đánh máy tính nên kéo dài hơi nhầm bạn ơi chứ không phải sai đề :))
a) \(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(A=\sqrt{\left(2+\sqrt{3}\right)\left(\sqrt{2+\sqrt{3}}+2\right)\left(-\sqrt{2+\sqrt{3}}+2\right)}\)
\(A=\sqrt{1}\)
\(A=1\)
b)\(B=\left(\frac{\sqrt{x}}{\sqrt{xy}-y}-\frac{\sqrt{y}}{\sqrt{xy}-x}\right).\left(x\sqrt{y}-y\sqrt{x}\right)\)
\(B=\frac{\sqrt{xy}}{\sqrt{xy}-y}x\sqrt{y}+\frac{\sqrt{x}}{\sqrt{xy}-y}y\sqrt{x}+\left(-\frac{\sqrt{y}}{\sqrt{xy}-x}\right)^2x\sqrt{y}+y\sqrt{x}\)
\(B=x\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{y}+y\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{x}+x\frac{\sqrt{x}}{\sqrt{xy}-x}\sqrt{y}-y\sqrt{x}\frac{\sqrt{y}}{\sqrt{xy}-y}\)
\(B=\frac{-x^{\frac{5}{2}}\sqrt{y}+\sqrt{x}.y^{\frac{5}{2}}}{\left(\sqrt{xy}-y\right)\left(\sqrt{xy}-x\right)}\)
\(B=\frac{\left(\sqrt{x}.y^{\frac{5}{2}}-x^{\frac{5}{2}}\sqrt{y}\right)\left(y+\sqrt{xy}\right)\left(x+\sqrt{xy}\right)}{\left(-y^2+xy\right)\left(-x^2+xy\right)}\)
c) \(C=\sqrt{\left(3-\sqrt{5}\right)^2+\sqrt{6}-2\sqrt{5}}\)
\(C=14-6\sqrt{5}+\sqrt{6}-2\sqrt{5}\)
\(C=14-8\sqrt{5}+\sqrt{6}\)
\(C=\sqrt{14-8\sqrt{5}+\sqrt{6}}\)