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Ta có công thức tổng quát: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*)

Áp dụng (*), ta được: \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+...+\left(\sqrt{100}-\sqrt{99}\right)=\sqrt{100}-\sqrt{1}=9\left(đpcm\right)\)

Trục căn thức ở mẫu : 

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{99}-\sqrt{100}\right)}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(=\sqrt{100}-\sqrt{1}\)

\(=10-1=9\)

=> đpcm

\(\sqrt{x^2+5x+20}=4\)

ĐK : x ∈ R

Bình phương hai vế

<=> \(x^2+5x+20=16\)

<=> \(x^2+5x+20-16=0\)

<=> \(x^2+5x+4=0\)(1)

Ta có : a - b + c = 1 - 5 + 4 = 0

=> (1) có hai nghiệm \(\hept{\begin{cases}x_1=-1\\x_2=-\frac{c}{a}=-\frac{4}{1}=-4\end{cases}}\)

\(\sqrt{x^2+5x+20}=4\)

\(\Rightarrow\left(\sqrt{x^2+5x+20}\right)^2=16\)

\(\Leftrightarrow x^2+5x+20=16\)

\(\Leftrightarrow x^2+5x+4=0\)

\(\Leftrightarrow\left(x^2+x\right)+\left(4x+4\right)=0\)

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-4\right\}\)

\(ĐKXĐ:y\ge0\)

\(x^2-4x+y-6\sqrt{y}+13=0\)

\(\Leftrightarrow x^2-4x+4+y-6\sqrt{y}+9=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

Vì \(\left(x-2\right)^2\ge0\forall x\); \(\left(\sqrt{y}-3\right)^2\ge0\forall y\ge0\)

\(\Rightarrow\left(x-2\right)^2+\left(\sqrt{y}-3\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=0\\\sqrt{y}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\\sqrt{y}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=9\left(t/m\right)\end{cases}}\)( t/m là thỏa mãn )

Vậy \(x=2\)và \(y=9\)

\(x^2-4x+y-6\sqrt{y}+13=0\)

<=> \(\left(x^2-4x+4\right)+\left(y-6\sqrt{y}+9\right)=0\)

<=> \(\left(x-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

Ta có : \(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\\left(\sqrt{y}-3\right)^2\ge0\forall y\ge0\end{cases}}\Leftrightarrow\left(x-2\right)^2+\left(\sqrt{y}-3\right)^2\ge0\forall x,\left(y\ge0\right)\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\\sqrt{y}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=9\left(tm\right)\end{cases}}\)

\(A=x^3+y^3-2xy=\left(x+y\right)\left(x^2-xy+y^2\right)-2xy=2x^2-4xy-2y^2\)

\(=2\left(x-y\right)^2\ge0.\text{ Dấu bằng: }x=y=1\)

a) \(ĐKXĐ:x\ge-1\)

\(\sqrt{x+1}=2\)\(\Rightarrow\left(\sqrt{x+1}\right)^2=4\)

\(\Rightarrow x+1=4\)\(\Leftrightarrow x=3\)( thỏa mãn ĐKXĐ )

Vậy \(x=3\)

b) \(ĐKXĐ:x\ge2\)

\(2\sqrt{x-2}< 6\)\(\Leftrightarrow\sqrt{x-2}< 3\)

Vì \(\sqrt{x-2}\ge0\); \(3>0\)

\(\Rightarrow\left(\sqrt{x-2}\right)^2< 9\)\(\Leftrightarrow x-2< 9\)

\(\Leftrightarrow x< 11\)

Kết hợp với ĐKXĐ \(\Rightarrow2\le x< 11\)

Vậy \(2\le x< 11\)

c) \(ĐKXĐ:x\ge4\)

 \(\sqrt{x^2-16}=-\sqrt{x-4}\)

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

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

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

Vì \(\sqrt{x+4}>0\)\(\Rightarrow\sqrt{x+4}+1>0\)

\(\Rightarrow\sqrt{x-4}=0\)\(\Leftrightarrow x-4=0\)\(\Leftrightarrow x=4\)

Vậy \(x=4\)

Ta có:

\(P=5x+4y+\frac{8}{x}+\frac{9}{y}\)

\(P=\left(\frac{8}{x}+2x\right)+\left(\frac{9}{y}+y\right)+3\left(x+y\right)\)

Áp dụng BĐT Cauchy ta được:

\(P\ge2\sqrt{\frac{8}{x}\cdot2x}+2\sqrt{\frac{9}{y}\cdot y}+3\cdot5\)

\(=2\cdot4+2\cdot3+15=29\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Vậy Min(P) = 29 khi \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Cảm ơn ạ

Áp dụng BĐT Bunyakovsky ta được:

\(\left(x+y\right)\left(\frac{2020}{x}+\frac{1}{2020y}\right)\ge\left(\sqrt{x}\cdot\sqrt{\frac{2020}{x}}+\sqrt{y}\cdot\sqrt{\frac{1}{2020y}}\right)\)

\(=\left(\sqrt{2020}+\sqrt{\frac{1}{2020}}\right)^2=2020+\frac{1}{2020}+2=2022\frac{1}{2020}\)

\(\Leftrightarrow\frac{2021}{2020}\cdot S\ge2022\frac{1}{2020}\)

\(\Rightarrow S\ge2022\frac{1}{2020}\div\frac{2021}{2020}=2021\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{\sqrt{x}}{\sqrt{\frac{2020}{x}}}=\frac{\sqrt{y}}{\sqrt{\frac{1}{2020y}}}\\x+y=\frac{2021}{2020}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020y\\x+y=\frac{2021}{2020}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

Vậy Min(S) = 2021 khi \(\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

Ta có: 

Vì \(\frac{2}{3}< x< \frac{13}{2}\Rightarrow\hept{\begin{cases}3x-2>0\\10-x>0\\13-2x>0\end{cases}}\)

Khi đó: \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\)

\(=\frac{1}{3x-2}+\frac{1}{10-x}+\frac{1}{13-2x}\) \(\left(1\right)\)

Áp dụng BĐT Cauchy Schwarz ta được:

\(\left(1\right)\ge\frac{\left(1+1+1\right)^2}{3x-2+10-x+13-2x}\)

\(=\frac{3^2}{21}=\frac{3}{7}\)

Vậy với \(\frac{2}{3}< x< \frac{13}{2}\) thì \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\ge\frac{3}{7}\)