Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) P = \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
P = \(\left(\frac{\sqrt{a}.\sqrt{a}-1}{2\sqrt{a}}\right)^2\cdot\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
P = \(\frac{\left(a-1\right)^2}{4a}\cdot\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
P = \(\frac{a-1}{4\sqrt{a}^2}\cdot\left(-4\sqrt{a}\right)\)
P = \(\frac{1-a}{\sqrt{a}}\)
b) với x > 0 và x khác 1
P < 0 => \(\frac{1-a}{\sqrt{a}}< 0\)
Do \(\sqrt{a}>0\) => 1 - a < 0 => a > 1
Vậy S = {a|a > 1}
Có 1 kiểu hơi khác Conan 1 tí -.-
\(a)P=\left(\frac{\sqrt{a}.\sqrt{a}-1}{2\sqrt{a}}\right).\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2.\frac{a-2\sqrt{a}+1-a-2\sqrt{1}-1}{a-1}=\frac{\left(a-1\right)\left(-4\sqrt{a}\right)}{\left(2\sqrt{a}\right)^2}\)
\(=\frac{\left(1-a\right).4\sqrt{a}}{4a}=\frac{1-a}{\sqrt{a}}\)
Vậy \(P=\frac{1-a}{\sqrt{a}}\)với a > 0 và \(a\ne1\)
b) Do a > 0 và a khác 1 nên P < 0 khi và chỉ khi :
\(\frac{1-a}{\sqrt{a}}< 0\Leftrightarrow1-a< 0\Leftrightarrow a>1\)
\(M=\left(\frac{x-\sqrt{x}+2}{x-1}-\frac{1}{\sqrt{x}-1}\right)\cdot\frac{x+2\sqrt{x}+1}{2x-2\sqrt{x}}\)
\(=\frac{\left(x-\sqrt{x}+2\right)-\sqrt{x}-1}{x-1}\cdot\frac{\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{x-1}\cdot\frac{\sqrt{x}+1}{2\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{2\sqrt{x}}\)
b) PT có nghiệm <=> x>0
<=>\(\sqrt{x}>0\)
<=> \(\sqrt{x}-1>-1\)
<=> x>-1
a) ĐKXĐ : \(a>0;a\ne1\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right)\)
\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)
\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)
b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
\(A=\left(\frac{1}{x-\sqrt{x}}\right)\div\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{x-2\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}}\)
Tại \(x=4+2\sqrt{3}\): \(\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(A=\frac{\sqrt{3}}{4+2\sqrt{3}+\sqrt{3}+1}=\frac{\sqrt{3}}{5+3\sqrt{3}}\)
\(A-1=\frac{\sqrt{x}-1}{x+\sqrt{x}}-1=\frac{-1-x}{x+\sqrt{x}}< 0\)do \(x>0\).
Vậy \(A< 1\).
Bài 1:
ĐK: $a\geq 0; a\neq 1$
a)
\(P=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)
\(=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)\)
\(=(\sqrt{a}+1)^2(\sqrt{a}-1)^2=(a-1)^2\)
b) \(P< 7-4\sqrt{3}\)
\(\Leftrightarrow (a-1)^2< (2-\sqrt{3})^2\)
\(\Leftrightarrow \sqrt{3}-2< a-1< 2-\sqrt{3}\)
\(\Leftrightarrow \sqrt{3}-1< a< 3-\sqrt{3}\)
Vậy $\sqrt{3}-1< a< 3-\sqrt{3}$ và $a\neq 1$
Bài 2:
a)
\(A=\frac{2}{a-\sqrt{a}}.\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}=\frac{2(\sqrt{a}-1)^2}{\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}=\frac{2(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}+1)}\)
b)
Xét hiệu \(A-1=\frac{2\sqrt{a}-2-a-\sqrt{a}}{\sqrt{a}(\sqrt{a}+1)}=-\frac{a-\sqrt{a}+2}{\sqrt{a}(\sqrt{a}+1)}\)
Thấy rằng: \(a-\sqrt{a}+2=(\sqrt{a}-\frac{1}{2})^2+\frac{7}{4}>0; \sqrt{a}(\sqrt{a}+1)>0 \) với mọi $a>0; a\neq 1$ nên:
\(A-1=-\frac{a-\sqrt{a}+2}{\sqrt{a}(\sqrt{a}+1)}<0\Rightarrow A< 1\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
a,Với \(a>0;a\ne1\)
\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{\sqrt{a}-1+a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{a-1}{a+\sqrt{a}}\)
b, Ta có : \(1=\frac{a+\sqrt{a}}{a+\sqrt{a}}\)mà \(a-1=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\)
\(a+\sqrt{a}=\sqrt{a}\left(\sqrt{a}+1\right)\)vì \(\sqrt{a}-1< \sqrt{a}\)
Vậy \(\frac{a-1}{a+\sqrt{a}}< 1\)hay \(M< 1\)