\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)

\(a,B=\dfrac{-\sqrt{x}-3+\sqrt{x}-3+x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\left(x\ge0;x\ne9\right)\\ B=\dfrac{x-2}{x-9}=\dfrac{x-9+7}{x-9}=1+\dfrac{7}{x-9}\in Z\\ \Leftrightarrow x-9\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{2;8;11;16\right\}\)

Vậy giá trị x thỏa đề là \(x=2\)

\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)

\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)

GTNN=-36 tại x=0

-36 bạn nha

CHÚC BẠN HỌC GIỎI

trả lời

GTNN = -36 tại x = 0

học tốt!!!

1:

a: \(A=2+3\sqrt{x^2+1}>=3\cdot1+2=5\)

Dấu = xảy ra khi x=0

b: \(B=\sqrt{x+8}-7>=-7\)

Dấu = xảy ra khi x=-8

1:

ĐKXĐ: \(x\notin\left\{3;-2;1\right\}\)

 \(A=\left(\dfrac{x\left(x+2\right)-x+1}{\left(x-3\right)\left(x+2\right)}\right):\left(\dfrac{x\left(x-3\right)+5x+1}{\left(x+2\right)\left(x-3\right)}\right)\)

\(=\dfrac{x^2+2x-x+1}{\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(x+2\right)\left(x-3\right)}{x^2-3x+5x+1}\)

\(=\dfrac{x^2+x+1}{\left(x-1\right)^2}\)

(x-1) (x+2) (x+3) (x+6) 
= [(x-1) (x+6)] . [(x+2) (x+3)]
=(x^2 +5x -6) (x^2+5x+6) 
=(x^2+5x)^2 - 6^2 = (x^2+5x)^2 - 36 
Vì (x^2+5x)^2 > hoặc bằng 0 => (x-1)(x+2)(x+3)(x+6) > hoặc bằng - 36. 
Dấu bằng xảy ra khi (x^2+5x)^2=0 <=> x=0 hoặc x=-5

(x - 1)(x + 2)(x + 3)(x + 6) 
= [(x - 1)(x + 6)].[(x + 2)(x + 3)] 
= (x² + 5x - 6)(x² + 5x + 6) 
= (x² + 5x)² - 36 >= -36 
=> min = -36 <=> x = 0 hoặc x = -5

#Đức Lộc#