Rút gọn
B=\(\sqrt{7+4\sqrt{3}}\)-\(2\sqrt{3}\)
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AH
Akai Haruma
Giáo viên
28 tháng 10 2021
Lời giải:
a. ĐKXĐ: $x>0; x\neq 4$
\(M=\frac{x}{\sqrt{x}(\sqrt{x}-2)}-\frac{4\sqrt{x}-4}{\sqrt{x}(\sqrt{x}-2)}=\frac{x-(4\sqrt{x}-4)}{\sqrt{x}(\sqrt{x}-2)}=\frac{x-4\sqrt{x}+4}{\sqrt{x}(\sqrt{x}-2)}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}(\sqrt{x}-2)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)
b.
\(x=3+2\sqrt{2}=(\sqrt{2}+1)^2\Rightarrow \sqrt{x}=\sqrt{2}+1\)
\(M=\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{2}+1-2}{\sqrt{2}+1}=3-2\sqrt{2}\)
c.
$M>0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}}>0$
$\Leftrightarrow \sqrt{x}-2>0$
$\Leftrightarrow \sqrt{x}>2\Leftrightarrow x>4$
Kết hợp đkxđ suy ra $x>4$
AA
6 tháng 8 2021
a, A= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}+x}{\left(\sqrt{x}+2\right)}\right)\)
A=\(\frac{1}{x+2\sqrt{x}}\)
b, A >= \(\frac{1}{3\sqrt{x}}\)
=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)
=> x <= -1 , x >= 4 (x khác 0)
16 tháng 5 2021
√x√x−2−6√x−4x−4(x\(\ge\)0,x\(\ne\)4)
=\(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}\)-\(\dfrac{6\sqrt{x}-4}{x-4}\)=\(\dfrac{x+2\sqrt{x}}{x-4}\)-\(\dfrac{6\sqrt{x}-4}{x-4}\)
=\(\dfrac{x+2\sqrt{x}-6\sqrt{x}+4}{x-4}\)=\(\dfrac{x-4\sqrt{x}+4}{x-4}\)=\(\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)(1)
b, với x=6-4\(\sqrt{2}\)=(2-\(\sqrt{2}\))^2 thay vào (1) ta được
\(\dfrac{\sqrt{\left(2-\sqrt{2}\right)}^2-2}{\sqrt{\left(2-\sqrt{2}\right)}^2+2}\)=\(\dfrac{2-\sqrt{2}-2}{2-\sqrt{2}+2}\)=\(\dfrac{-\sqrt{2}}{4-\sqrt{2}}\)=\(\dfrac{\sqrt{2}}{\sqrt{2}-4}\)
16 tháng 5 2021
a)ĐKXĐ: x≠4;x≥0
=\(\dfrac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
=\(\dfrac{x+2\sqrt{x}-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b) thế x=\(6-4\sqrt{2}\) (thỏa mãn) vào bt ta đc:
\(\dfrac{\sqrt{6-4\sqrt{2}}-2}{\sqrt{6-4\sqrt{2}}+2}\)=\(\dfrac{\sqrt{\left(2-\sqrt{2}\right)^2}-2}{\sqrt{\left(2-\sqrt{2}\right)^2}+2}\)=\(\dfrac{-\sqrt{2}}{4-\sqrt{2}}\)=\(\dfrac{-1}{\sqrt{2}-1}\)=\(-\sqrt{2}-1\)
29 tháng 6 2021
b, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Ta có : \(B=\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4-x+2\sqrt{x}-4+x+2}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+2}{\sqrt{x}}\)
29 tháng 6 2021
b) Ta có: \(B=\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+x+2}{\sqrt{x}}\)
c) Ta có: \(C=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3-5+\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)
\(=2+\sqrt{3}-2\sqrt{3}\)
\(=2-\sqrt{3}\)