-(5*căn bậc hai(2)*căn bậc hai(3)-17*căn bậc hai(2)-5)/(5*căn bậc hai(2))

\(\frac{2}{5-\sqrt{3}}+\frac{5}{2\sqrt{2}+3}-\sqrt{8}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{22}+\frac{5\left(2\sqrt{2}-3\right)}{-1}-\sqrt{8}\)

\(=\frac{5+\sqrt{3}}{11}-10\sqrt{2}+15-2\sqrt{2}\)

\(=\frac{5+\sqrt{3}}{11}-12\sqrt{2}+15=\frac{5+\sqrt{3}}{11}-\frac{11\left(12\sqrt{2}-15\right)}{11}\)

\(=\frac{5+\sqrt{3}-132\sqrt{2}+165}{11}=\frac{\sqrt{3}-132\sqrt{2}+170}{11}\)

\(\frac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3-2\sqrt{6}+2}}{\sqrt{3}-\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=1\)

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\(\sqrt{2x+8}=0\)ĐK : \(x\ge-4\)

\(\Leftrightarrow2x+8=0\Leftrightarrow x=-4\)

\(\sqrt{x-4\sqrt{x-4}}=2\left(ĐK:x\ge4\right)\)

\(\Leftrightarrow\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}-2\right)^2}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}-2\right|=2\)(1)

Với \(4\le x< 8\)(1) <=> \(2-\sqrt{x-4}=2\Leftrightarrow\sqrt{x-4}=0\Leftrightarrow x=4\left(tm\right)\)

Với \(x\ge8\)(1) <=> \(\sqrt{x-4}-2=2\Leftrightarrow\sqrt{x-4}=4\Leftrightarrow x-4=16\Leftrightarrow x=20\left(tm\right)\)

Vậy S = { 4 ; 20 }

Thay m=0 vào hàm số ta có 

\(y=f\left(x\right)=\left(0^2-0+1\right)x-3.0=x\)x 

Cho x=0 => y=0

Cho x=1 =>y=1 ta đc điểm (1;1)

Cho x=2 => y=2 ta đc điểm (2;2)

Vẽ đg thằng đi qua gốc toạ độ và đi qua điểm (1;1) và (2;2) ta đc đồ thị hàm số y=x

P  +3 = \(\left(\frac{x+1}{2x+5}+1\right)+\left(\frac{x+2}{2x+4}+1\right)+\left(\frac{x+3}{2x+3}+1\right)=\frac{3x+6}{2x+5}+\frac{3x+6}{2x+4}+\frac{3x+6}{2x+3}\)

\(=\left(3x+6\right)\left(\frac{1}{2x+5}+\frac{1}{2x+4}+\frac{1}{2x+3}\right)\)

Ta chứng minh \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

<=> (a + b + c)(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)) \(\ge\)9 

=> \(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)

<=> \(1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+1\ge9\)

<=> \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)(0)

Lại có \(\frac{b}{a}+\frac{a}{b}\ge2\)(1) 

Thật vậy \(\frac{b^2+a^2}{ab}\ge2\)

<=> (b - a)² \(\ge\)0 (đúng) 

Tương tự được \(\frac{c}{a}+\frac{a}{c}\ge2;\frac{b}{c}+\frac{c}{b}\ge2\) (2) 

Từ (1) (2) => (0) đúng => ĐPCM

Dấu "=" xảy ra <=> a = b = c

Khi đó P + 3 \(\ge\left(3x+6\right)\left(\frac{9}{2x+5+2x+4+2x+3}\right)=\frac{9}{2}\)

=> P \(\ge\frac{3}{2}\left(ĐPCM\right)\)

\(x\ge3\)

học tốt

xin tiick

ĐKXĐ : x² - 9 ≥ 0 <=> ( x - 3 )( x + 3 ) ≥ 0

<=> x ≥ 3 hoặc x ≤ -3 

Để phân thức có nghĩa khi 

 \(\frac{-3}{1-5x}\ge0\Rightarrow1-5x< 0\Rightarrow-5x< -1\Leftrightarrow x>\frac{1}{5}\)