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27 tháng 11 2024

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

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

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

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

\(=\dfrac{x\sqrt{x}+3x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt[]{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

b: Thay \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) vào P, ta được:

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

7 tháng 12 2024

a) ĐKXĐ: 0<x\(\ne\)1

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

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

    P=\(\dfrac{x\sqrt{x}+2x}{x\sqrt{x}+x-2\sqrt{x}}\)+\(\dfrac{2\sqrt{x}-2+x+2}{x\sqrt{x}+x-2\sqrt{x}}\)

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

    P=\(\dfrac{x\sqrt{x}+2x+2\sqrt{x}+x}{x\sqrt{x}+x-2\sqrt{x}}\)

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

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

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

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

b) ĐKXĐ: 0<x\(\ne\)1

     x=3+2\(\sqrt{2}\)

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

     x=(\(\sqrt{2}\)+1)2

     Thay x=3+2\(\sqrt{2}\)(TM)vào P:

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

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

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

       P=1+\(\sqrt{2}\)

Vậy P=1+\(\sqrt{2}\) khi x=3+2\(\sqrt{2}\)

18 tháng 5 2018

Câu c mk ko piết làm. Bạn Thoòng cảm

18 tháng 5 2018

Hàm số bậc nhất

AH
Akai Haruma
Giáo viên
24 tháng 7 2018

Lời giải:

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

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

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

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

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

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

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

b) \(P-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{2\sqrt{x}-(x+1)}{3(x+\sqrt{x}+1)}\)

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

Với \(x\neq 1, x\geq 0\Rightarrow -(\sqrt{x}-1)^2< 0; x+\sqrt{x}+1>0\)

Do đó: \(P-\frac{1}{3}< 0\Rightarrow P< \frac{1}{3}\)

Bài 2:

a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)

\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)

\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)

b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

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

17 tháng 7 2018

\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)

\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)

\(\Leftrightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow x< 4\)

Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)

KL............

\(2.\) Tương tự bài 1.

\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)

30 tháng 6 2018

có phải/....

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

\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)

2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

30 tháng 6 2018

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

AH
Akai Haruma
Giáo viên
14 tháng 5 2018

Lời giải:

a)

Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)

Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)

\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)

\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)

\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)

b)

ĐK: \(x\geq 0; x\neq 4\)

\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)

\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)

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

 

 

AH
Akai Haruma
Giáo viên
14 tháng 5 2018

c) Thêm ĐK: \(x\geq 1\)

Từ biểu thức P vừa tìm được:

\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

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

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

\((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)

\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :

\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)

Vậy..........

1 tháng 2 2019

\(a)A=\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\dfrac{2+\sqrt{8}}{1+\sqrt{2}}\\ A=\dfrac{\left(\sqrt{3}-\sqrt{6}\right)\left(1+\sqrt{2}\right)}{1^2-\left(\sqrt{2}\right)^2}-\dfrac{\left(2+\sqrt{8}\right)\left(1-\sqrt{2}\right)}{1^2-\left(\sqrt{2}\right)^2}\\ A=-\left(\sqrt{3}+\sqrt{6}-\sqrt{6}-2\sqrt{3}\right)+2-2\sqrt{2}+2\sqrt{2}-4\\ A=\sqrt{3}-2\)

\(b)B=\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\\ B=\left[\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\\ B=\dfrac{\sqrt{x}+2-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\left(\sqrt{x}+2\right)\\ B=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}+2\right)\\ B=\dfrac{4}{x-4}\)

Bài 1: 

a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)

b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)

=>3 căn x=3

=>căn x=1

hay x=1(loại)

8 tháng 8 2018

1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)

8 tháng 8 2018

Làm nốt nè :3

\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{x-2}{2x}>0\)

\(\Leftrightarrow x-2>0\left(do:x>0\right)\)

\(\Leftrightarrow x>2\)

\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)

\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)

Kết hợp với DKXĐ : \(0< a< 1\)

Bài 2: 

a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)

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

b: Để P>=-2 thì P+2>=0

\(\Leftrightarrow-2\sqrt{a}+2>=0\)

=>0<=a<1