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1 tháng 10 2020

\(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a+\sqrt{ab}+b\)

1 tháng 10 2020

Thiếu đkxđ: \(a,b>0\)

a: \(P=-5\sqrt{\dfrac{160}{90}}=-5\cdot\dfrac{4}{3}=-\dfrac{20}{3}\)

b: \(Q=\sqrt{a}-\sqrt{b}+2\sqrt{b}=\sqrt{a}+\sqrt{b}\)

22 tháng 7 2017

Sửa đề: GTLN

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{a}{a+\sqrt{2019a+bc}}=\frac{a}{a+\sqrt{a\left(a+b+c\right)+bc}}=\frac{a}{a+\sqrt{a^2+ab+ca+bc}}\)

\(=\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}\)

\(=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{b}{b+\sqrt{2019b+ac}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}};\frac{c}{c+\sqrt{2019c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng theo vế 3 BĐT trên ta có:

\(P\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

4 tháng 10 2020

\(A=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)

ĐKXĐ : x > 1

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

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

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

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

\(=\frac{x}{\sqrt{x}-1}\)

Để A = 9/2

=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )

<=> 2x = 9( √x - 1 )

<=> 2x = 9√x - 9

<=> 2x + 9 = 9√x (1)

Bình phương hai vế

(1) <=> 4x2 + 36x + 81 = 81x

     <=> 4x2 + 36x + 81 - 81x = 0

     <=> 4x2 - 45x + 81 = 0

     <=> 4x2 - 36x - 9x + 81 = 0

     <=> 4x( x - 9 ) - 9( x - 9 ) = 0

     <=> ( x - 9 )( 4x - 9 ) = 0

     <=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )

a).  \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)

\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)

\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)

\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)

\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\) 

ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC

9 tháng 8 2016

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

Biểu thức  \(A\)  có nghĩa khi  \(\hept{\begin{cases}\sqrt{x}+1\ne0;\text{ }x\ge0\\\sqrt{x}-1\ne0\end{cases}}\)  \(\Leftrightarrow\)  \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có:

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

\(A=\frac{x+\sqrt{x}-2\sqrt{x}+2-2\sqrt{x}-2}{x-1}=\frac{x-3\sqrt{x}}{x-1}\)

Vậy,  \(A=\frac{x-3\sqrt{x}}{x-1}\)

9 tháng 8 2016

đề đúng hk bn

1 tháng 9 2020

a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)

\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}+\sqrt{2}\right)\)

\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)

\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)

b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{1}\right)\)

\(=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)

\(=2\sqrt{5+2\sqrt{5}+1}.\left(\sqrt{5}-1\right)\)

\(=2\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\)

\(=2.\left(5-1\right)=2.4=8\)