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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)
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
\(a.Q=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}}{x+1}\right)\)
\(ĐKXĐ:x\) ≥ \(0;x\) # \(1\)
\(Q=\left(\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{x-\sqrt{x}+1}{x+1}\)
\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(b.\) Ta thấy : \(x-\sqrt{x}+1=x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
Mà : \(\sqrt{x}+1>0\)
⇒ \(Q>0\)
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)
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: \(x\geq 0; x\neq 4\)
\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)
\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)
\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c) Thêm ĐK: \(x\geq 1\)
Từ biểu thức P vừa tìm được:
\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)
Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)
\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). 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..........
a: Sửa đề: \(Q=\left(\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\dfrac{3-\sqrt{x}}{x-1}\)
\(=\dfrac{x+\sqrt{x}+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để Q=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>x=1(loại)
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)\(A=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}=\dfrac{x+2}{\sqrt{x}^3-1^3}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x+2+x-1-1}{\sqrt{x}^3-1}=\dfrac{2x}{\sqrt{x}^3-1}\)
Ta có: \(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\cdot\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)