Ta có:A²=\(\dfrac{9x^2-30x+25}{1-x^2}\)

=>\(A^2-A^2x^2=9x^2-30x+25\)

<=>(9+A²)x²-30x+25-A²=0(1)

Với -1<x<1 thì A luôn xác định khi đó (1) sẽ luôn có nghiệm

=>\(\Delta'\ge0\)

<=>15²-(9+A²).(25-A²)\(\ge\)0

<=>225+A⁴-16A²-225\(\ge\)0

<=>A²(A²-16)\(\ge\)0

<=>A²\(\ge\)16 hoặc 0\(\ge\)A²

<=>4\(\ge\)A\(\ge\)-4 hoặc A=0

Nhưng:A=0 thì 5-3x=0<=>x=\(\dfrac{5}{3}\)(L)

=>(1) có nghiệm <=>4\(\ge\)A\(\ge\)-4

=>GTNN của A=-4 khi và chỉ khi \(\Delta'=0\)<=>x=\(\dfrac{15}{9+A^2}=\dfrac{15}{9+16}=\dfrac{15}{25}=\dfrac{3}{5}\)(TMĐKXĐ)

Vậy...

\(1\)\(<\)\(x<-1\)\(\Rightarrow\)\(5-3x>0\)\(\Rightarrow y>0\)

Nhân chéo 2 vế ta được:

\(y^2=\frac{\left(5-3x\right)^2}{1-x^2}\)\(\Rightarrow-x^2y^2+y^2=25-30x+9x^2\)

\(\Leftrightarrow x^2.\left(9+y^2\right)-30x+25-y^2=0\)(1)

\(\Delta'=15^2-\left(25-y^2\right)\left(9+y^2\right)\Leftrightarrow\Delta=y^4-16y^2\)

Để ý có GTNN thì phương trình (1) phải có nghiệm

\(\Rightarrow\Delta\ge0\Leftrightarrow y^2.\left(y^2-16\right)\ge0\Rightarrow y^2\ge16\)

\(\Leftrightarrow y\ge4\left(TM\right)\)hoac \(y\le-4\left(KTM\right)\)

Vay \(y\ge4\)khi\(x=\frac{15}{25}\)

y²= (5-3x)²/ ( 1-x²)

y²= ( 25+9x²-30x) / ( 1-x²)

y² = (  16-16 x² +25x²-30x+9) / ( 1-x²)

y² = 16 + (5x-3)² / ( 1-x²)

vì -1<x<1 => x²<1 => 1-x²>0

=> ( 5x-3)²/ (1-x²) >= 0

=> y²>=16

=> y>= 4 => min y =4 

dấu = xảy ra <=> x=5/3

# Bài 1

* Ta cm BĐT sau \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\) (1) bằng cách biến đổi tương đương

* Với \(x,y>0\) áp dụng (1) ta có

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\left(\sqrt{x}\right)^2}+\dfrac{1}{\left(\sqrt{y}\right)^2}\ge\dfrac{1}{2}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\)

Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\)

\(\Rightarrow\) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\le1\) \(\Leftrightarrow\) \(0< \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\le1\) (I)

* Ta cm BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (2)

Áp dụng (2) với x , y > 0 ta có

\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge\dfrac{4}{\sqrt{x}+\sqrt{y}}\) (II)

* Từ (I) và (II) \(\Rightarrow\) \(\dfrac{4}{\sqrt{x}+\sqrt{y}}\le1\)

\(\Leftrightarrow\) \(\sqrt{x}+\sqrt{y}\ge4\)

Dấu "=" xra khi \(x=y=4\)

Vậy min \(\sqrt{x}+\sqrt{y}=4\) khi \(x=y=4\)

1.ĐK:\(x\ge0,x\ne9\)

\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}-3}\)

\(=\left[\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right].\dfrac{\sqrt{x}-3}{\sqrt{x}-5}\)

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

Để \(P< \dfrac{-1}{2}\Leftrightarrow\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-5\right)}< \dfrac{-1}{2}\)

Câu 3

a, ĐKXĐ: x>0, x\(\ne\)4

M=( \(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\)). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b, Thay x= \(6+4\sqrt{2}\) ( x>0, x\(\ne\)4) ta có:

M= \(\dfrac{\sqrt{6+4\sqrt{2}}}{\sqrt{6+4\sqrt{2}}-2}\)

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

= \(\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)}{\sqrt{2}}\) = \(1+\sqrt{2}\)

Vậy khi x= \(6+4\sqrt{2}\) thì M= \(1+\sqrt{2}\)

c, Để M<1 <=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 1\)

<=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)

<=> \(\dfrac{2}{\sqrt{x}-2}< 0\)

Vì 2>0 <=> \(\sqrt{x}-2< 0\)

<=> \(\sqrt{x}< 2\)

<=> x<4

Vậy để M<1 thì 0<x<4

<=>

Câu 2

a, \(\sqrt{3x+2}=5\) (x\(\ge\dfrac{-2}{3}\))

<=> \(\sqrt{3x+2}=\sqrt{25}\)

<=> 3x+2=25

<=> 3x= 23

<=> x=\(\dfrac{23}{3}\)

Vậy S= \(\left\{\dfrac{23}{3}\right\}\)

2.

a/ Áp dụgn hệ quả bđt cô si,ta có :

\(A=xy+yz+zx\le\dfrac{\left(x+y+z\right)}{3}=\dfrac{a^2}{3}\)

Vậy GTLN A =a^2/3 khi x= y =z =a/3

b/Áp dụng BĐT Cô-Si dạng Engel,ta có :

\(B=\dfrac{x^2}{1}+\dfrac{y^2}{1}+\dfrac{z^2}{z}\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{a^2}{3}\)

Vậy GTNN của B = a^2/2 khi x=y=z =a/3

\(B=\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+7\ge2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}+7=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)

Vậy min B = \(\left(2+\sqrt{3}\right)^2\) khi \(\dfrac{3x}{1-x}=\dfrac{4\left(1-x\right)}{x}\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)

a) Để biểu thức P xác định thì \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

Vậy ĐKXĐ:x\(\ge0\),x\(\ne9\)

\(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\left(-3\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)

b) Ta có \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{1}{2}\Leftrightarrow-6< \sqrt{x}+3\Leftrightarrow\sqrt{x}>-9\)

Vì \(\sqrt{x}\ge0\) và 0>-9

Vậy \(x\ge0\)

Kết hợp với ĐKXĐ, Vậy \(x\ge0\) và \(x\ne9\) thì P<\(\dfrac{1}{2}\)