Ta có:

\(A=\sqrt{1-x}+\sqrt{1+x}\) \(\left(-1\le x\le1\right)\)

\(=1.\sqrt{1-x}+1.\sqrt{1+x}\)

Áp dụng BĐT Bunhiacopxki, ta có:

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

\(\le\sqrt{\left(1^2+1^2\right).\left(1-x+1+x\right)}=\sqrt{2.2}=2\)

Vậy \(A_{max}=2\), đạt được khi và chỉ khi \(\dfrac{1}{\sqrt{1-x}}=\dfrac{1}{\sqrt{1+x}}\Leftrightarrow1-x=1+x\Leftrightarrow x=0\)

Bài 5: 

a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:

\(E=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1-3}=\dfrac{\sqrt{3}}{\sqrt{3}-2}=-3-2\sqrt{3}\)

b: Để E<1 thì E-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

c: Để E nguyên thì \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)

hay \(x\in\left\{16;25;49\right\}\)

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

\(a,B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\\ B=x-\sqrt{x}+1-\sqrt{x}=\left(\sqrt{x}-1\right)^2\)

Mà \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)

a: \(B=\left(\sqrt{x}-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

b: \(A=\dfrac{2x+1-x+\sqrt{x}}{x\sqrt{x}-1}\cdot\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)

a:

ĐKXĐ: x>=0; x<>1

 Sửa đề: \(M=x-\dfrac{2x-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)

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

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

b: \(M=x-\sqrt{x}+2\)

\(=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>x=1/4

1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)

Thay \(x=\frac{1}{9}\) vào A ta có:

\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)

2. \(B=...\)

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

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

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

3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{\sqrt{x}+3}{-6}\)

Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)

hay \(P\le-\frac{1}{2}\)

Dấu "=" xảy ra <=> x=0

toán lớp 9 khó zậy em đọc k hỉu 1 phân số

a. ĐKXĐ: \(x>0\)

\(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{x+\sqrt{x}}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

b. Để \(P=-1\) thÌ  \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=-1\) 

\(\Leftrightarrow x+\sqrt{x}+1=-\sqrt{x}\)

\(\Leftrightarrow x+2\sqrt{x}+1=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\)

\(\Leftrightarrow\sqrt{x}+1=0\)

\(\Leftrightarrow\sqrt{x}=-1\) ( vô lý )

Vậy không có x thỏa mãn ycbt

c. Ta có \(x=\dfrac{8}{\sqrt{5}-1}-\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8-8\sqrt{5}+8}{5-1}=\dfrac{16}{4}=4\)

Thay x=4 vào P, ta được

\(P=\dfrac{4+\sqrt{4}+1}{\sqrt{4}}=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)

d. \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\) \(\Rightarrow P-3=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-3\)

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

Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-1\right)^2\ge0\\\sqrt{x}>0\end{matrix}\right.\) \(\Rightarrow P-3\ge0\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

Vậy \(P_{min}=3\) khi \(x=1\)

a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)

\(\Rightarrow A=\frac{2+1}{2+2}=\frac{3}{4}\)

Vậy với x = 4 thì A = 3/4 

b, \(B=\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}+5}{x-1}=\frac{3\left(\sqrt{x}+1\right)-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

với x=4(t/m DK)

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

thay\(\sqrt{x}\)=2 vào biểu thức A ta được

A=(2+1)/(2+2)

A=3/4

Đặt   \(A=\sqrt{x+8}+\sqrt{x-1}\)

Xét   \(A^2=\left|x+8\right|+2\sqrt{\left(x+8\right)\left(x-1\right)}+\left|x-1\right|\)

       \(\Leftrightarrow A^2=x+8+1-x+2\sqrt{\left(x+8\right)\left(x-1\right)}\)

        \(\Leftrightarrow A^2=9+2\sqrt{\left(x+8\right)\left(x-1\right)}\)

Ta có  \(A^2\ge9\)\(\Rightarrow\)min A = 3  \(\Leftrightarrow x=1\)

nhỏ nhất là 9 khi x= -8 hoặc x=1