\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)

\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)

\(=\frac{1-\sqrt{25}}{-1}=4\)

\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)

\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)

\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)

\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)

\(=1\)

\(C=\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}: \frac{1}{x^2-\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}.(x^2-\sqrt{x})\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}.\sqrt{x}(\sqrt{x^3}-1)\)

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

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

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

\(\sqrt{x}=1\Leftrightarrow x=1\\ \sqrt{x}=3\Leftrightarrow x=9\\ \sqrt{x}=5\Leftrightarrow x=25\\ \sqrt{x}=7\Leftrightarrow x=49\\ \sqrt{x}=9\Leftrightarrow x=81\\ \sqrt{x+1}=11\\ \Leftrightarrow x+1=121\\ \Leftrightarrow x=120\)

a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)

c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9

`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)

\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)

\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

`b,`Ta có :

 \(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)

\(M=\dfrac{3\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(3+\sqrt{2}\right)^2}-3}\)

\(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}\)

\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)

\(M=\dfrac{6+9\sqrt{2}}{2}\)

`c,`  Để `M<1` Ta có :

 \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)

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

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

\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )

\(\sqrt{x}< 3\)

\(x< 9\)

Đối chiếu ĐKXĐ ta có : `0<x<9`

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

xog xét 2 TH

b, bình phương 

2

GTLN : 2 dấu = xra \(2\le x\le4\)

Hà Thị Thế pạn làm ra lun giúp mjk dx k ạ

Lời giải:

Ta có \(x+y=-\sqrt{3}; xy=\frac{1}{2}\)

\(x^{11}+y^{11}=(x^5+y^5)(x^6+y^6)-x^5y^5(x+y)=(x^5+y^5)(x^6+y^6)+\frac{\sqrt{3}}{32}\)

Nhận thấy:

\(x^2+y^2=(x+y)^2-2xy=3-2.\frac{1}{2}=2\)

\(x^3+y^3=(x+y)^3-3xy(x+y)=-3\sqrt{3}+\frac{3\sqrt{3}}{2}=-1,5\sqrt{3}\)

\(x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)\)

\(=-3\sqrt{3}+\frac{1}{4}\sqrt{3}=\frac{-11}{4}\sqrt{3}\)

\(x^6+y^6=(x^3+y^3)^2-2(xy)^3=(-1,5\sqrt{3})^2-2.\frac{1}{8}=\frac{13}{2}\)

Do đó: \(x^{11}+y^{11}=\frac{-11}{4}\sqrt{3}.\frac{13}{2}+\frac{\sqrt{3}}{32}=\frac{-571}{32}\sqrt{3}\)

Ta có : x+y= -1 và xy= \(\dfrac{-1}{2}\)

x²+y²= (x+y)²-2xy=1-1=0

x⁴+y⁴ = (x²+y²)²-2x²y²=0+\(\dfrac{1}{2}\)=\(\dfrac{1}{2}\)

x⁸+y⁸=(x⁴+y⁴)²-2x⁴y⁴=\(\dfrac{1}{4}\)-\(\dfrac{1}{8}\)=\(\dfrac{1}{8}\)

x³+y³=(x+y)³-3xy(x+y)=-1\(-\dfrac{3}{2}\) =\(\dfrac{-1}{2}\)

x¹¹+y¹¹=(x⁸+y⁸)(x³+y³)-x³y³(x⁵+y⁵)=\(\dfrac{1}{8}\).\(\dfrac{-1}{2}\)+\(\dfrac{1}{8}\)(x⁵+y⁵) 

Bạn tính x⁵+y⁵ rồi thế vô ( Tính x³+y³ và x²+y² rồi làm giống cách trên chứ dài quá mình viết không nổi  )

Mình sửa lại đề tí:

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

ĐKXĐ: \(x\ge1\)

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

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

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

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

TH1: \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\) pt trở thành:

\(\sqrt{x-1}-\left(\sqrt{x-1}-1\right)=1\) (luôn đúng)

TH2: \(1\le x< 2\)

\(\Rightarrow\sqrt{x-1}-\left(1-\sqrt{x-1}\right)=1\)

\(\Leftrightarrow2\sqrt{x-1}=2\Rightarrow x=2\) (ktm)

Vậy nghiệm của pt là \(x\ge2\)

\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\) (ĐK: \(x\ne1;x\ge0\))

\(A=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

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

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

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

\(A=\dfrac{1}{\sqrt{x}+2}\)