Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B4
a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b)\(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}=\sqrt{5}+\sqrt{2}\)
c)\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{\left(\sqrt{2}+1\right)^2}{1}=\left(\sqrt{2}+1\right)^2\)
d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
B3
a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
-\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x\right|-6x+9\)
\(x< 0\)
\(--->x+3-x-6x+9\)
\(=\left(x-x\right)-6x+3+9\)
\(=-6x+\left(3+9\right)=-6x+12\)
\(x>0\)
\(--->3+x+x-6x+9\)
\(=\left(x+x-6x\right)+\left(3+9\right)\)
\(=\left(2x-6x\right)+12\)
\(=4x+12\)
a) Ta có: \(x=9\)thỏa mãn đk
\(\Rightarrow\)Thay \(x=9\)vào biểu thức ta được:
\(A=\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=\frac{-9}{2}\)
b) Với x thỏa mãn ĐKXĐ thì ta có:
\(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}\)
\(=\frac{1}{\sqrt{x}+2}+\frac{x+14}{x-4}-\frac{4}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)+\left(x+12\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2+x+12-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
1. x = 9 => A = \(\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=-\frac{9}{2}\)
2. \(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}=\frac{\sqrt{x}-2+x+12-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
3. \(AB>-\frac{3}{4}\) <=> \(\frac{3\sqrt{x}}{1-\sqrt{x}}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+2}>-\frac{3}{4}\)
<=> \(-\frac{3\sqrt{x}}{\sqrt{x}+2}+\frac{3}{4}>0\)
<=> \(\frac{12\sqrt{x}-3\sqrt{x}-4}{4\left(\sqrt{x}+2\right)}< 0\)
<=> \(\frac{9\sqrt{x}-4}{4\sqrt{x}+8}< 0\)
Do \(4\sqrt{x}+8>0\)với mọi x => \(9\sqrt{x}-4< 0\) <=> \(x< \frac{16}{81}\)
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{-\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x\sqrt{x}+x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\dfrac{-2x^2+x\sqrt{x}-2\sqrt{x}+1+2x^2-x\sqrt{x}-2x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{-2x-\sqrt{x}+1}\)
\(=\dfrac{-\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{-\sqrt{x}\left(2x+\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
b: Thay \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) vào A, ta được:
\(A=\dfrac{17-12\sqrt{2}-\sqrt{2}+1+1}{3-2\sqrt{2}}=\dfrac{19-13\sqrt{2}}{3-2\sqrt{2}}=5-\sqrt{2}\)