Δ = b² - 4ac = [ -( m - 1 ) ]² + 12

= ( m - 1 )² + 12 ≥ 12 > 0 ∀ m

hay phương trình luôn có hai nghiệm phân biệt ∀ m ( đpcm )

a, \(P=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\)

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

\(=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x\sqrt{x}-24-3x+8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{x+8}{\sqrt{x}+1}\)

b, Ta co : \(x=14-6\sqrt{5}=14-2.3.\sqrt{5}\)

\(=3-2.3\sqrt{5}+\left(\sqrt{5}\right)^2=\left(3-\sqrt{5}\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)

Thay vào P ta được : 

\(P=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{2\left(29-\sqrt{5}\right)}{11}\)

\(\hept{\begin{cases}x^4-x^3+3x^2-4y-1=0\left(1\right)\\\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{x^2+2xy+4y^2}{3}=x+2y}\end{cases}}\)

Ta có: \(\frac{x^2+4y^2}{2}=\frac{2\left(x^2+4y^2\right)}{4}=\frac{\left(1^2+1^2\right)\left(x^2+4y^2\right)}{4}\ge\frac{\left(x+2y^2\right)}{4}\)( bunhiacopxki ) 

\(\Rightarrow\sqrt{\frac{x^2+4y^2}{2}}\ge\frac{x+2y}{2}\)

Lại có: \(\sqrt{\frac{x^2+2xy+4y^2}{3}}\ge\frac{x+2y}{2}\)( 2) 

Chứng minh: Đẳng thức cần chứng minh \(\Leftrightarrow\frac{x^2+2xy+4y^2}{3}\ge\frac{x^2+4xy+4y^2}{4}\)

\(\Leftrightarrow4x^2+8xy+16y^2\ge3x^2+12xy+12y^2\)

\(\Leftrightarrow x^2-4xy+4y^2\ge0\)

\(\Leftrightarrow\left(x-2y\right)^2\ge0;\forall x,y\)

=> (2) đúng )

\(\Rightarrow\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{x^2+2xy+4y^2}{3}}\ge x+2y\)

Dấu "=" xảy ra \(\Leftrightarrow x=2y\)

Thay x=2y vào (1) ta được: 

\(x^4-x^3+3x^2-2x-1=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\Rightarrow y=\frac{1}{2}\\x^2+3x+1=0\left(3\right)\end{cases}}\)

\(\Delta_{\left(3\right)}=3^2-4=5>0\)

\(\Rightarrow\)pt (3) có 2 nghiệm phân biệt \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\Rightarrow y=\frac{-3+\sqrt{5}}{4}\\x=\frac{-3-\sqrt{5}}{2}\Rightarrow y=\frac{-3-\sqrt{5}}{4}\end{cases}}\)

Vậy ... 

a)

b)

c)

d)

 a)

b)

c)

d)

Giải:

#Học tốt!!!

a) 7 /24

b) 1, 08

c) 17/2

d) 15 /29

Ta có: \(y^2+xy+x+y+5=y^2+xy+x+y+4+1\)

\(=y^2+xy+x+y+\left(x+y\right)\left(x+1\right)+1\)

\(=\left(x+y+1\right)^2\)

\(x^3+y^3+12y+13=x^3+y^3+12\left(y+1\right)+1\)

\(=x^3+y^3+3\left(x+y\right)\left(x+1\right)\left(y+1\right)+1\)

\(=x^3+y^3+3\left(x+y\right)\left(xy+x+y+1\right)+1\)

\(=x^3+y^3+3xy\left(x+y\right)+3\left(x+y\right)\left(x+y+1\right)+1\)

\(=\left(x+y+1\right)^3\)

Khi đó phương trình thứ hai tương đương với

\(\left(x+y+1\right)^5=243\Leftrightarrow x+y+1=3\)

Từ đây kết hợp phương trình một ta được \(x=y=1\).

a, Ta có  \(\sqrt{25-16}=\sqrt{9}=3\)

\(\sqrt{25}-\sqrt{16}=5-4=1\)

Do 3 > 1 nên \(\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)

a) căn 25 - 16  > căn 25 - căn 16

b)Với $a>b>0$ nên  \sqrt{a},\sqrt{b},\sqrt{a-b}a​,b​,− đều xác định

Để so sánh \sqrt{a}-\sqrt{b}a​−b​ và $-$ ta quy về so sánh \sqrt{a}a​ và \sqrt{a-b}+\sqrt{b}−+b​.

+) (\sqrt{a})^2=a(a​)2=a.

+) (\sqrt{a-b}+\sqrt{b})^2=(\sqrt{a-b})^2+2\sqrt{a-b}.\sqrt{b}+(\sqrt{b})^2=a-b+b+2\sqrt{a-b}.\sqrt{b}=a+2\sqrt{a-b}.\sqrt{b}(−+b​)2=(−)2+2−.b​+(b​)2=a−b+b+2−.b​=a+2−

.b​.

Do $a>b>0$ nên 2\sqrt{a-b}.\sqrt{b}>02−.b​>0

$\Rightarrow$ a+2\sqrt{a-b}.\sqrt{b}>aa+2−.b​>a

$\Rightarrow$ (\sqrt{a-b}+\sqrt{b})^2>(\sqrt{a})^2(−+b​)2>(a​)2

Do \sqrt{a},\sqrt{a-b}+\sqrt{b}>0a​,−+b​>0 

$\Rightarrow$ \sqrt{a-b}+\sqrt{b}>\sqrt{a}−+b​>a​

$\iff$ \sqrt{a-b}>\sqrt{a}-\sqrt{b}−>a​−b​ (đpcm)

Vậy \sqrt{a-b}>\sqrt{a}-\sqrt{b}−>a​−b​.