Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
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 ...
Bài 2 :
b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)
ĐKXĐ : \(x\ge1\)
Pt(1) tương đương :
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)
Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)
\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\) ( Thỏa mãn )
Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)
Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\Leftrightarrow2=2\) ( Luôn đúng )
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)
Bài 1 :
a) ĐKXĐ : \(-1\le a\le1\)
Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)
\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)
\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)
\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)
Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)
b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :
\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)
Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)