a: \(P=\dfrac{3}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-5}{x-1}\)

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

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

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

b: \(x=24-16\sqrt{2}=8\left(3-2\sqrt{2}\right)=8\left(\sqrt{2}-1\right)^2\)

Thay \(x=8\left(\sqrt{2}-1\right)^2\) vào P, ta được:

\(P=\dfrac{1}{\sqrt{8\left(\sqrt{2}-1\right)^2}-1}\)

\(=\dfrac{1}{2\sqrt{2}\left(\sqrt{2}-1\right)-1}=\dfrac{1}{4-2\sqrt{2}-1}\)

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

a: \(Q=\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)

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

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

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

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

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

b: P=Q

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

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

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

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

mà \(\sqrt{x}-1>=-1\) với mọi x thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1=\sqrt{3}\)

=>\(\sqrt{x}=1+\sqrt{3}\)

=>\(x=\left(1+\sqrt{3}\right)^2=4+2\sqrt{3}\left(nhận\right)\)

a: \(A=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt[]{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}+3}=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)

\(=\dfrac{4x}{\sqrt{x}-3}\)

b: A=-2

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

=>\(4x+2\sqrt{x}-6=0\)

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

=>\(\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

mà \(2\sqrt{x}+3>=3>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1=0\)

=>x=1(nhận)

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

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-\left(a-4\right)}\)

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

b: P>1/6

=>P-1/6>0

=>\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)

=>\(\dfrac{6\left(\sqrt{a}-2\right)-3\sqrt{a}}{18\sqrt{a}}>0\)

=>\(6\left(\sqrt{a}-2\right)-3\sqrt{a}>0\)

=>\(3\sqrt{a}-12>0\)

=>\(\sqrt{a}>4\)
=>a>16

1: Thay x=9 vào A, ta được:

\(A=\dfrac{3\cdot3}{3+2}=\dfrac{9}{5}\)

2: \(B=\dfrac{x+4}{x-4}-\dfrac{2}{\sqrt{x}-2}\)

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

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

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

3: \(A-B< \dfrac{3}{2}\)

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

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

=>\(\dfrac{4\sqrt{x}-3\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+2\right)}< 0\)

=>\(\dfrac{\sqrt[]{x}-6}{2\left(\sqrt{x}+2\right)}< 0\)

=>\(\sqrt{x}-6< 0\)

=>\(\sqrt{x}< 6\)

=>0<=x<36

mà x là số nguyên dương lớn nhất thỏa mãn

nên x=35

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

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

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

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

a: \(\left(\sqrt{\dfrac{4}{3}}+\sqrt{3}\right)\cdot\sqrt{6}\)

\(=\sqrt{\dfrac{4}{3}\cdot6}+\sqrt{3\cdot6}\)

\(=\sqrt{8}+\sqrt{18}=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}\)

b: \(\left(1-2\sqrt{5}\right)^2=\left(2\sqrt{5}-1\right)^2\)

\(=\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot1+1\)

\(=21-4\sqrt{5}\)

c: \(2\sqrt{3}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)

d: \(\sqrt{45}-\sqrt{20}+\sqrt{5}\)

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

\(=4\sqrt{5}-2\sqrt{5}=2\sqrt{5}\)

a). 5\(\sqrt2\)

b). 21-4\(\sqrt5\)

c). -\(\sqrt3\)

d). 2

\(\dfrac{1}{a^3+b^3+abc}=\dfrac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)+abc}\le\dfrac{1}{\left(a+b\right)\left(2ab-ab\right)+abc}=\dfrac{1}{ab\left(a+b\right)+abc}=\dfrac{1}{ab\left(a+b+c\right)}\)
tương tự với các hạng tử còn lại, ta được
\(Vetrai\le\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\left(\dfrac{1}{a+b+c}\right)=\dfrac{a+b+c}{abc}\cdot\dfrac{1}{a+b+c}=\dfrac{1}{abc}\)
dấu bằng xảy ra khi a=b=c

àm lại nếu tôi làm sai

Và làm tiếp tục giúp tôi

1 D

2 C

3 B

4 A

5 A

6 C

7 B

8 A

9 A

10 B

Chúc mừng mọi người nhé! Mọi người giỏi ghê 

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