Cho biểu thức \(Q=\frac{4}{\sqrt{b}-1}+\frac{3}{\sqrt{b}+1}-\frac{6\sqrt{b}+2}{b-1}.\left(b\ge0;b\ne1\right).\)
- Rút gọn\(Q\)
- Tính giá trị của biểu thức\(Q\)khi \(b=6+2\sqrt{5}\)
K
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DV
0
Những câu hỏi liên quan
1 tháng 5 2019
B=\(\left(\frac{\sqrt{b}\left(\sqrt{b}-2\right)}{\left(\sqrt{b}+2\right)\left(\sqrt{b}-2\right)}-\frac{\sqrt{b}\left(\sqrt{b}+2\right)}{\left(\sqrt{b}+2\right)\left(\sqrt{b}-2\right)}+\frac{4\sqrt{b}-1}{b-4}\right)\)
B=\(\frac{\left(b-2\sqrt{b}\right)-\left(b+2\sqrt{b}\right)+\left(4+\sqrt{b}-1\right)}{\left(\sqrt{b}+2\right)\left(\sqrt{b-2}\right)}\)
B=\(\frac{b-2\sqrt{b}-b-2\sqrt{b}+4+\sqrt{b}-1}{\left(\sqrt{b}+2\right)\left(\sqrt{b}-2\right)}\)
B=
3 tháng 6 2019
Bài 2
P=a-(\(\frac{1}{\sqrt{a}-\sqrt{a-1}}-\frac{1}{\sqrt{a}+\sqrt{a-1}}\)
P=a-(\(\frac{\sqrt{a}+\sqrt{a-1}}{a-a+1}-\frac{\sqrt{a}-\sqrt{a-1}}{a-a+1}\)
P=a-\(2\sqrt{a-1}\)
P=a-1-2\(\sqrt{a-1}+1\)
P=\(\left(\sqrt{a-1}-1\right)^2\)
Có \(\left(\sqrt{a-1}-1\right)^2>=0vớimọix\)
=> P >=0
AH
Akai Haruma
Giáo viên
2 tháng 6 2020
Lời giải:
a)
\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)
b)
\(B=\left(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x}=\frac{x-1}{x}\)
Y
6 tháng 7 2019
a.
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Vậy với \(0\le x< 9;x\ne1\) thì ..........
6 tháng 7 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)
Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)
6 tháng 7 2019
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x-2\sqrt{x}+1}{x-1}\)
\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)
ND
23 tháng 10 2019
a, x = \(\frac{4\left(\sqrt{3}+1\right)}{3-1}-\frac{4\left(\sqrt{3}-1\right)}{3-1}\)
x = \(\left(2\sqrt{3}+2\right)-\left(2\sqrt{3}-2\right)\)
x = \(2\sqrt{3}+2-2\sqrt{3}+2\)
x = 4 (TMĐK)
=> A = \(\frac{2\sqrt{4}+1}{3\sqrt{4}+1}\)
=> A = \(\frac{5}{7}\)
Vậy x = \(\frac{4}{\sqrt{3}-1}-\frac{4}{\sqrt{3}+1}\) thì A = \(\frac{5}{7}\)
b, B = \(\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
B = \(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\)
B = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
c, \(\frac{B}{A}>2\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}:\frac{2\sqrt{x}+1}{3\sqrt{x}+1}\) > 2
<=> \(\frac{3\sqrt{x}+1}{\sqrt{x}+1}>2\)
<=> \(\frac{3\sqrt{x}+1}{\sqrt{x}+1}-2>0\)
<=> \(\frac{3\sqrt{x}+1-2\sqrt{x}-2}{\sqrt{x}+1}>0\)
<=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
mà \(\sqrt{x}+1>0\) \(\forall\) \(x\in\) ĐKXĐ
=> \(\sqrt{x}-1>0\)
<=> \(\sqrt{x}>1\)
<=> \(x>1\)
Kết hợp ĐKXĐ : x \(\ge0\) ; x \(\ne\) 1
=> x > 1 thì \(\frac{B}{A}>2\)