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`a,`
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\\ =\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{4}{\sqrt{x}+1}\)
`b,` Để `A *B<0` ta có :
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{4}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.4>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)
Kết hợp với đkxđ ta có : \(0< x< 1\)
a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Thay x=3+2\(\sqrt{2}\)
A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0
\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)
a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
a: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
b: Để A<=3/căn x thì \(\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}< =\dfrac{3}{\sqrt{x}}\)
=>\(\dfrac{x-2\sqrt{x}-1-3x+6\sqrt{x}-3}{\left(\sqrt{x}-1\right)^2}< =0\)
=>\(-2x+4\sqrt{x}-4< =0\)
=>\(x-2\sqrt{x}+2>=0\)(luôn đúng)
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
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 \(x\in\left\{0;1;2;3\right\}\)
1.\(x=4\)
\(B=\left(\dfrac{x+1}{2}-\sqrt{x}\right)=\left(\dfrac{4+1}{2}-\sqrt{4}\right)=\dfrac{5}{2}--2=\dfrac{5-4}{2}=\dfrac{1}{2}\)
2.\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)=\left(\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+1}{2}-\sqrt{x}=\dfrac{x+1-2\sqrt{x}}{2}=\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(M=A.B=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
3.\(M=\dfrac{\sqrt{x}}{6}\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{6}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=6\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}=6\sqrt{x}-6\)
\(\Leftrightarrow x-5\sqrt{x}+6=0\)
Đặt \(\sqrt{x}=a;a\ge0\)
=> pt trở thành:
\(a^2-5a+6=0\)
\(\Delta=\left(-5\right)^2-4.6=25=24=1>0\)
=> pt có 2 nghiệm:
\(\left\{{}\begin{matrix}x_1=\dfrac{5+\sqrt{1}}{2}=3\left(tm\right)\\x_2=\dfrac{5-\sqrt{1}}{2}=2\left(tm\right)\end{matrix}\right.\)
Xét \(\sqrt{a}=3\)
\(\Leftrightarrow a=9\)
Xét \(\sqrt{a}=2\)
\(\Leftrightarrow a=4\)
Vậy \(x=9;4\)
a: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{4}{\sqrt{x}+1}\)
b: Để A=-B thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-4}{\sqrt{x}+1}\)
=>\(\left(\sqrt{x}+1\right)^2=-4\left(\sqrt{x}-1\right)\)
=>\(x+2\sqrt{x}+1+4\sqrt{x}-4=0\)
=>\(x+6\sqrt{x}-3=0\)
=>\(x+6\sqrt{x}+9-12=0\)
=>\(\left(\sqrt{x}+3\right)^2=12\)
=>\(\left[{}\begin{matrix}\sqrt{x}+3=2\sqrt{3}\\\sqrt{x}+3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\sqrt{3}-3\left(vôlý\right)\\\sqrt{x}=2\sqrt{3}-3\end{matrix}\right.\)
=>\(\sqrt{x}=2\sqrt{3}-3\)
=>\(x=\left(2\sqrt{3}-3\right)^2=21-12\sqrt{3}\)