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Lời giải:
a)
Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)
Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)
\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)
\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)
b)
ĐK: \(x\geq 0; x\neq 4\)
\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)
\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)
\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c) Thêm ĐK: \(x\geq 1\)
Từ biểu thức P vừa tìm được:
\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)
Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)
\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :
\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)
Vậy..........
a/ Sai đề.
\(x+2\sqrt{2x-4}=\left(x-2\right)+2.\sqrt{2}.\sqrt{x-2}+2=\left(\sqrt{2}+\sqrt{x-2}\right)^2\)
b/ \(M=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{\left(\sqrt{2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{x-2}\right)^2}\)
\(=\sqrt{2}+\sqrt{x-2}+\left|\sqrt{2}-\sqrt{x-2}\right|\)
1. Nếu \(2\le x\le4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\)
2. Nếu \(x>4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
-\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x\right|-6x+9\)
\(x< 0\)
\(--->x+3-x-6x+9\)
\(=\left(x-x\right)-6x+3+9\)
\(=-6x+\left(3+9\right)=-6x+12\)
\(x>0\)
\(--->3+x+x-6x+9\)
\(=\left(x+x-6x\right)+\left(3+9\right)\)
\(=\left(2x-6x\right)+12\)
\(=4x+12\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
\(\sqrt{x+2\sqrt{x+1}}\)
\(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(\left|\sqrt{x-1}+1\right|\)
a ) Với \(x\ge1\) ta có :
\(M=\sqrt{x+2\sqrt{x-1}}=\sqrt{x-1+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}+1\right)^2}=\sqrt{x-1}+1\)
b ) Với \(x\ge\frac{1}{2}\) ta có : \(N=\sqrt{2x-1+4\sqrt{2x-1}+4}=\sqrt{\left(\sqrt{2x-1}+2\right)^2}=\sqrt{2x-1}+2\)
\(\sqrt{2x+3+4\sqrt{2x-1}}\)
\(=\sqrt{\sqrt{\left(2x-1+2\right)^2}}\)
\(=\left|2x-1+2\right|\)
M=\(\sqrt{x-1+1}\) N=\(\sqrt{2x-1+2}\)
a) M =\(\sqrt{x-1}+1\)
b) N= \(\sqrt{2x-1}+2\)
a) M =\(\sqrt{X-1}+1\)
B) N = \(\sqrt{2x-1}+2\)
a) M=\(\sqrt{x-1}+1\)
b) N= \(\sqrt{2x-1}+2\)
\(a,M=\sqrt{x+2\sqrt{x-1}}vớix\ge1\\ =\sqrt{x-1+2\sqrt{x-1}+1}\\ =\sqrt{\left(\sqrt{x-1}+1\right)^2}=\sqrt{x-1}+1\left(TMDKXD\right)\\ b,vớix\ge\dfrac{1}{2}tqcó\\ N=\sqrt{2x+3+4\sqrt{2x-1}}\\ =\sqrt{\left(\sqrt{2x}-1+2\right)^2}=\sqrt{2x-1}+2\)
a\()\)M=\(\sqrt{x-1}+1\)
b\()\)N=\(\sqrt{2x-1}+2\)
a)M=\(\sqrt{x+2\sqrt{x-1}}\left(x\le1\right)\)
=\(\sqrt{(x-1+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}+1\right)^2}=\left|\sqrt{x-1}+1\right|\)
a) M= \(\sqrt{x-1}+1\)b) N= \(\sqrt{2x-1}+2\)
a) M =\(\sqrt{x-1}+1\)
b) N =\(\sqrt{2x-1}+2\)
a, M=\(\sqrt{x-1+1}\)
b,N=\(\sqrt{2x-1+2}\)
a) M=\sqrt{x-1}+1M=x−1+1.
b) N=\(\sqrt{2x-1}\)+ 2
\(\sqrt{2x-1}\).
a) M = \(\sqrt{x-1}\)+ 1
b) N = \(\sqrt{2x-1}\)+ 2
a) M = \(\sqrt{x-1}\)+ 1
b) N = \(\sqrt{2x-1}\)+ 2
a)M=\(\sqrt{x-1}+1\)
b)N=\(\sqrt{2x-1}+2\)
a = căn (x - 1) + 1
b =
a) \(\sqrt{x-1}\)+1
b) \(\sqrt{2x-1}\)+2
b) N=\(\sqrt{2x-1}\)+2
a) M=\(\sqrt{x-1}\)+1
a) M = \(\sqrt{x-1}\)+1
b) N = \(\sqrt{2x-1}\)+ 2
M= \(\sqrt{x+1}\)+ 1
N=\(\sqrt{2x-2}\)+ 2
a) M = \(\sqrt{x-1}\)+ 1
b) N = \(\sqrt{2x-1}\)+ 2
a)M=\(\sqrt{x-1}+1\)
b)N=\(\sqrt{2x-1}+2\)
a,M=\(\sqrt{x+2\sqrt{x-1}}\)=\(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)=\(\sqrt{x-1}+1\)
b,N\(\sqrt{2x+3+4\sqrt{2x-1}}\)=\(\sqrt{\left(\sqrt{2x-1}+2\right)^2}\)=\(\sqrt{2x-1}+2\)
a) M = \(\sqrt{x-1}+1\)
b) N = \(\sqrt{2x-1}+2\)
a) M=\(\sqrt{x+2\sqrt{x-1}}\)= \(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}\)=\(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)=\(|\sqrt{x-1}+1|\)
có:\(\sqrt{x-1}\)≥0⇒\(\sqrt{x-1}+1\)≥ 1>0
⇒M=\(\sqrt{x-1}+1\)
b) N=\(\sqrt{2x+3+4\sqrt{2x-1}}=\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+4}\)=\(\sqrt{\left(\sqrt{2x-1}+2\right)^2}\)=\(|\sqrt{2x-1}+2|\)
có\(\sqrt{2x-1}\ge0\)⇒ \(\sqrt{2x-1}+2\ge2>0\)
⇒N=\(\sqrt{2x+1}+2\)
a) \(M=\sqrt{x+2\sqrt{x-1}}=\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x+1}+1\right)^2}=|\sqrt{x+1}+1|\)
Có: \(\sqrt{x +1}\ge0\Rightarrow\sqrt{x+1}+1\ge1>0\)
\(\Rightarrow M=\sqrt{x-1}+1\)
b) \(N=\sqrt{2x+3+4\sqrt{2x-1}}=\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+4}=\sqrt{\left(\sqrt{2x-1}+2\right)^2}=|\sqrt{2x-1}+2|\)
Có: \(\sqrt{2x-1}>0\Rightarrow\sqrt{2x-1}+2\ge2>0\)
\(\Rightarrow N=\sqrt{2x-1}+2\)
a) M= \(\sqrt{x-1}\)+ 1
b) N= \(\sqrt{2x-1}\)+2