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ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)
= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{x+2}{\sqrt{x}-1}\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999
\(a)A=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\sqrt{5}+3+\sqrt{3}-\left(\sqrt{5}+3\right)\\ =\sqrt{3}\)
\(b)B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2\\ =\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\\ =2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\\ =2\left(25-21\right)=8\)
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 = \(\dfrac{1}{4}\)
\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .