mik cần gấp ạ^^

a: Khi x=25 thì \(A=\dfrac{5+1}{5-2}=\dfrac{6}{3}=2\)

b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1-\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+4}{x-\sqrt{x}-2}\)

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

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

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

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

c: P=B:A

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

P<-1

=>P+1<0

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

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

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

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

a) Thay x=64 vào Q ta có:

\(Q=\dfrac{\sqrt{64}-2}{\sqrt{64}-3}=\dfrac{8-2}{8-3}=\dfrac{6}{5}\)

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

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

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

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

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

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

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

\(=4-2\sqrt{3}+2\sqrt{3}\)

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

A=-3; B=

1) Thay x=1x=1 vào biểu thức: A=√1+2√1−2A=1+21−2
A=−3A=−3
2) Chứng minh B=√x√x+2B=xx+2 với x≥0,x≠4x≥0,x≠4.
B=√x+2(√x−2)(√x+2)+(√x+1)(√x−2)(√x+2)(√x−2)−2√x(√x+2)(√x−2)B=x+2(x−2)(x+2)+(x+1)(x−2)(x+2)(x−2)−2x(x+2)(x−2)
=√x+2+x−√x−2−2√x(√x+2)(√x−2)=x−2√x(√x+2)(√x−2)=x+2+x−x−2−2x(x+2)(x−2)=x−2x(x+2)(x−2)
=√x(√x−2)(√x+2)(√x−2)=√x√x+2=x(x−2)(x+2)(x−2)=xx+2
3) Tìm xx để A⋅B≥0A⋅B≥0
A⋅B=√x+2√x−2⋅√x√x+2=√x√x−2A⋅B=x+2x−2⋅xx+2=xx−2.
TH1: x=0⇒√x=0⇒A⋅B=0x=0⇒x=0⇒A⋅B=0 (TM)
TH2: x>0⇒√x>0⇒√x−2>0⇒x>4x>0⇒x>0⇒x−2>0⇒x>4
Kết hợp điêu kiện: x=0x=0 hoặc x>4x>4 thỏa mãn yêu cầu.

x

tick cho em la em lam lien

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

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

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

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

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

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

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

b) Mà với \(x\ge0\) và \(x\ne1\) thì 

\(x+\sqrt{x}+1\ge0\) và \(2>0\) nên \(P>0\)

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

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

b: x+căn x+1+1>=1>0

2>0

=>P>0 với mọi x thỏa mãn x>=0 và x<>1

1. Với x = 36
=> A= \(\dfrac{\sqrt{36}-2}{\sqrt{36}-1}\)=\(\dfrac{4}{5}\)
2. Với x >0, x ≠1
B=\(\dfrac{x-5}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}\)
B=\(\dfrac{x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{x-5-2\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{x-5-2\sqrt{x}+2+4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
3. P=\(\dfrac{A}{B}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\). \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Ta có \(\sqrt{P}< \dfrac{1}{2}\)
=>P<\(\dfrac{1}{4}\)
=> \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)<\(\dfrac{1}{4}\)
=> \(4\left(\sqrt{x}-2\right)< \sqrt{x}+1\)
=> \(4\sqrt{x}-8< \sqrt{x}+1 \)
=> \(3\sqrt{x}< 9\)
=>\(\sqrt{x}< 3\)
=> x< 9
Lại có x ϵ Z => x ϵ {-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8}
Ta thử lại với x ≠ 1
=> x ϵ {-8,-7,-6,-5,-4,-3,-2,0,2,3,4,5,6,7,8}

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

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

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

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

3: A/B>3/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)

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

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)

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

=>-căn x>-2

=>căn x<2

=>0<x<4

1) Thay x=64 vào A ta có:

\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)

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

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

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

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

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

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

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

3) Ta có:

\(\dfrac{A}{B}>\dfrac{3}{2}\) khi

\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

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

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

Mà: \(2\sqrt{x}\ge0\forall x\)

\(\Leftrightarrow2-\sqrt{x}>0\)

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

\(\Leftrightarrow x< 4\)

Kết hợp với đk:

\(0< x< 4\)